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Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | Well, it's going to be some large number. Actually, I know what it is. It's 256, because in the last video, we figured out that 2 to the eighth was equal to 256. And so 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9. What's 8 to the third? |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And so 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9. What's 8 to the third? It's 64 times, right? 8 times 8 squared is 64, so 8 cubed. So let's see. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | What's 8 to the third? It's 64 times, right? 8 times 8 squared is 64, so 8 cubed. So let's see. 4 times 2 is 3. 6 times 8, it looks like it's 512. Correct. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | So let's see. 4 times 2 is 3. 6 times 8, it looks like it's 512. Correct. And there's other ways you could have done it, because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right? |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | Correct. And there's other ways you could have done it, because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually, it's this exponent property where you can multiply. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually, it's this exponent property where you can multiply. When you take something to an exponent and then take that to an exponent, and you can essentially just multiply the exponents, that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation. There's a whole video on proving it a little bit more formally. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And actually, it's this exponent property where you can multiply. When you take something to an exponent and then take that to an exponent, and you can essentially just multiply the exponents, that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation. There's a whole video on proving it a little bit more formally. The next logarithm property I'm going to show you, and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | There's a whole video on proving it a little bit more formally. The next logarithm property I'm going to show you, and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are a calculator addict? Well, let's say you go to class and there's a quiz. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are a calculator addict? Well, let's say you go to class and there's a quiz. The teacher says, you can use your calculator. And using your calculator, I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator and not find it. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | Well, let's say you go to class and there's a quiz. The teacher says, you can use your calculator. And using your calculator, I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator and not find it. Because there is no log base 17 number, a button on your calculator. You'll probably either have a log button or you'll have an LN button. And just so you know, the log button on your calculator is probably base 10. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And you will scramble and look for the log base 17 button on your calculator and not find it. Because there is no log base 17 number, a button on your calculator. You'll probably either have a log button or you'll have an LN button. And just so you know, the log button on your calculator is probably base 10. And your LN number, your LN button on your calculator is going to be base E. For those of you who aren't familiar with E, don't worry about it. But it's 2.71 something something. It's a number. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And just so you know, the log button on your calculator is probably base 10. And your LN number, your LN button on your calculator is going to be base E. For those of you who aren't familiar with E, don't worry about it. But it's 2.71 something something. It's a number. It's nothing. It's an amazing number, but we'll talk more about that. In a future presentation. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | It's a number. It's nothing. It's an amazing number, but we'll talk more about that. In a future presentation. So there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | In a future presentation. So there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. We could do either E or 10. But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. We could do either E or 10. But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. Then you can clear it or if you know how to use a parenthesis on your calculator, you can do that. But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam. And then you just divide them and you get your answer. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. Then you can clear it or if you know how to use a parenthesis on your calculator, you can do that. But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam. And then you just divide them and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And then you just divide them and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job, so just so you know, the logarithms are useful. Let's do some examples. So let's just rewrite a bunch of things in simpler forms. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job, so just so you know, the logarithms are useful. Let's do some examples. So let's just rewrite a bunch of things in simpler forms. So if I wanted to write the log base 2 of the square root of 32 divided by the square root of 8, how can I rewrite this so it's reasonably not messy? Well, let's think about this. This is the same thing. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | So let's just rewrite a bunch of things in simpler forms. So if I wanted to write the log base 2 of the square root of 32 divided by the square root of 8, how can I rewrite this so it's reasonably not messy? Well, let's think about this. This is the same thing. This is equal to, I don't know if I move vertically or horizontally, but I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | This is the same thing. This is equal to, I don't know if I move vertically or horizontally, but I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing, and we learned that at the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing, and we learned that at the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm. Let's keep the 1 half out. That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm. Let's keep the 1 half out. That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? Let's see. Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8. We learned that property at the beginning of this presentation. |
Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3 | That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? Let's see. Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8. We learned that property at the beginning of this presentation. And then if we want, we can distribute this original 1 half, and this equals 1 half log base 2 of 32 minus 1 half, minus 1 fourth, because we have to distribute that 1 half, minus 1 fourth log base 2 of 8. This is 5 halves minus, this is 3, 3 times 1 fourth minus 3 fourths. Or 10 fourths minus 3 fourths is equal to 7 fourths. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | Use a calculator to find log base e of 67 to the nearest thousandth. So just as a reminder, e is one of these crazy numbers that shows up in nature and finance and all these things and it's approximately equal to 2.71 and it just keeps going on and on and on. So you could view log base e as 67. You might say, what does e mean? E is just a number, just like pi is just a number. This is really the same thing as saying log base 2.71 and the actual number. So you'd have to write all the digits that keep on going forever and never repeat of 67. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | You might say, what does e mean? E is just a number, just like pi is just a number. This is really the same thing as saying log base 2.71 and the actual number. So you'd have to write all the digits that keep on going forever and never repeat of 67. So what power do I have to raise e to to get to 67? So another way of saying that is if this is equal to x, you're saying e to the x is equal to 67. We need to figure out what x is. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | So you'd have to write all the digits that keep on going forever and never repeat of 67. So what power do I have to raise e to to get to 67? So another way of saying that is if this is equal to x, you're saying e to the x is equal to 67. We need to figure out what x is. Now, traditionally you will never see someone write log base e even though e is one of the most common bases to take a logarithm of. So the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. And I think that's used because e shows up so many times in nature. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | We need to figure out what x is. Now, traditionally you will never see someone write log base e even though e is one of the most common bases to take a logarithm of. So the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. And I think that's used because e shows up so many times in nature. So log base e of 67, another way of saying that or seeing that and the more typical way of seeing that is the natural log. And I think this is LN, so I think it's maybe from French or something, log natural of 67. So this is the same thing as log base e of 67. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | And I think that's used because e shows up so many times in nature. So log base e of 67, another way of saying that or seeing that and the more typical way of seeing that is the natural log. And I think this is LN, so I think it's maybe from French or something, log natural of 67. So this is the same thing as log base e of 67. This is saying the exact same thing. To what power do I have to raise e to to get 67? When you see this LN, it literally means log base e. Now, they let us use a calculator and that's good because I don't know off the top of my head what power I have to raise 2.71 and so on and so forth. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | So this is the same thing as log base e of 67. This is saying the exact same thing. To what power do I have to raise e to to get 67? When you see this LN, it literally means log base e. Now, they let us use a calculator and that's good because I don't know off the top of my head what power I have to raise 2.71 and so on and so forth. What power I have to raise that to to get to 67. So we'll get our calculator out. So we get the TI-85 out. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | When you see this LN, it literally means log base e. Now, they let us use a calculator and that's good because I don't know off the top of my head what power I have to raise 2.71 and so on and so forth. What power I have to raise that to to get to 67. So we'll get our calculator out. So we get the TI-85 out. And different calculators will have different ways of doing it. If you have a graphing calculator like this, you can literally type in the statement natural log of 67 and then evaluate it. So here this is the button for LN, means natural log, log natural maybe. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | So we get the TI-85 out. And different calculators will have different ways of doing it. If you have a graphing calculator like this, you can literally type in the statement natural log of 67 and then evaluate it. So here this is the button for LN, means natural log, log natural maybe. LN of 67 and then you press enter and it will give you the answer. If you don't have a graphing calculator, you might have to press 67 and then press natural log to give you the answer. But a graphing calculator can literally type it in the way that you would write it out. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | So here this is the button for LN, means natural log, log natural maybe. LN of 67 and then you press enter and it will give you the answer. If you don't have a graphing calculator, you might have to press 67 and then press natural log to give you the answer. But a graphing calculator can literally type it in the way that you would write it out. And then you would press enter. So 4.20469 and we want to round to the nearest thousandth. So this is a thousandths place right here, this 4. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | But a graphing calculator can literally type it in the way that you would write it out. And then you would press enter. So 4.20469 and we want to round to the nearest thousandth. So this is a thousandths place right here, this 4. The digit after that is 5 or larger, it's a 6, so we're going to round up. So this is 4.205. This is approximately equal to 4.205. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | So this is a thousandths place right here, this 4. The digit after that is 5 or larger, it's a 6, so we're going to round up. So this is 4.205. This is approximately equal to 4.205. And it actually makes a lot of sense because we know that E is greater than 2 and it is less than 3. And if you think about what 2 to the fourth power gets you to 16 and 3 to the fourth power gets you to 81. 67 is between 16 and 81 and E is between 2 and 3. |
Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3 | This is approximately equal to 4.205. And it actually makes a lot of sense because we know that E is greater than 2 and it is less than 3. And if you think about what 2 to the fourth power gets you to 16 and 3 to the fourth power gets you to 81. 67 is between 16 and 81 and E is between 2 and 3. So at least it feels right that something that's like 2.71 to the little over the fourth power should get you to a number that's pretty close to 3 to the fourth power. And actually that makes sense because it's actually closer to 3. 2.71 is closer to 3 than it is to 2. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | It says divide the polynomials. The form of your answer should either be a straight up polynomial or a polynomial plus the remainder over x minus five, which we have here in the denominator, where p of x is a polynomial and k is an integer. So we've done stuff like this, but like always, I encourage you to pause this video and work on this on your own. And if you were doing this on Khan Academy, there's a little bit of an input box here where you'd have to type in the answer. But let's just do it on paper for now. All right, so we're trying to figure out what x minus five divided into two x to the third power. Actually, I wanna be careful here because I wanna be very, very organized about my different degree columns. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | And if you were doing this on Khan Academy, there's a little bit of an input box here where you'd have to type in the answer. But let's just do it on paper for now. All right, so we're trying to figure out what x minus five divided into two x to the third power. Actually, I wanna be careful here because I wanna be very, very organized about my different degree columns. So this is my third degree column. And then I want my second degree column, but there is no second degree term here. There's a first degree term, so I'll write it out here. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | Actually, I wanna be careful here because I wanna be very, very organized about my different degree columns. So this is my third degree column. And then I want my second degree column, but there is no second degree term here. There's a first degree term, so I'll write it out here. So minus 47x, and actually, to be even more careful, I'll write plus zero x squared. And then I have minus 15. By putting that plus zero x squared, that's making sure I'm doing good, I guess, degree place column hygiene. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | There's a first degree term, so I'll write it out here. So minus 47x, and actually, to be even more careful, I'll write plus zero x squared. And then I have minus 15. By putting that plus zero x squared, that's making sure I'm doing good, I guess, degree place column hygiene. All right, so now we can work through this. And first, we could say, hey, how many times does x go into the highest degree term here? Well, x goes into two x to the third power, two x squared times. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | By putting that plus zero x squared, that's making sure I'm doing good, I guess, degree place column hygiene. All right, so now we can work through this. And first, we could say, hey, how many times does x go into the highest degree term here? Well, x goes into two x to the third power, two x squared times. And we'd wanna put that in the second degree column. Two x squared. You can see how it would've gotten messy if I put negative 47x here. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | Well, x goes into two x to the third power, two x squared times. And we'd wanna put that in the second degree column. Two x squared. You can see how it would've gotten messy if I put negative 47x here. I'd be like, where do I put that two x squared? And you might confuse yourself, which none of us would want to happen. All right, two x squared times negative five is negative 10x squared. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | You can see how it would've gotten messy if I put negative 47x here. I'd be like, where do I put that two x squared? And you might confuse yourself, which none of us would want to happen. All right, two x squared times negative five is negative 10x squared. Two x squared times x is two x to the third power. Now we wanna subtract what we have in red from what we have in blue. So I'll multiply them both by negative one, so that becomes a negative. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | All right, two x squared times negative five is negative 10x squared. Two x squared times x is two x to the third power. Now we wanna subtract what we have in red from what we have in blue. So I'll multiply them both by negative one, so that becomes a negative. And then that one becomes a positive. And that's actually one of the biggest areas for careless errors. If you have negative here, and you just wanna subtract it because you know you have to subtract, we're like, no, I'm subtracting a negative. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | So I'll multiply them both by negative one, so that becomes a negative. And then that one becomes a positive. And that's actually one of the biggest areas for careless errors. If you have negative here, and you just wanna subtract it because you know you have to subtract, we're like, no, I'm subtracting a negative. It needs to be a positive now. All right, so zero x squared plus 10x squared is 10x squared. And then the two x to the third minus two x to the third is just zero. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | If you have negative here, and you just wanna subtract it because you know you have to subtract, we're like, no, I'm subtracting a negative. It needs to be a positive now. All right, so zero x squared plus 10x squared is 10x squared. And then the two x to the third minus two x to the third is just zero. And then we can bring down that negative 47x. And once again, we look at the highest degree terms. X goes into 10x squared 10x times. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | And then the two x to the third minus two x to the third is just zero. And then we can bring down that negative 47x. And once again, we look at the highest degree terms. X goes into 10x squared 10x times. So plus 10x. 10x times negative five is negative 50x. Negative 50x. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | X goes into 10x squared 10x times. So plus 10x. 10x times negative five is negative 50x. Negative 50x. 10x times x is 10x squared. And once again, we wanna subtract what we have in teal from what we have in red. So we can multiply both of these times negative one. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | Negative 50x. 10x times x is 10x squared. And once again, we wanna subtract what we have in teal from what we have in red. So we can multiply both of these times negative one. That becomes a negative. This one becomes a positive. Now, negative 47x plus 50x is positive three x. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | So we can multiply both of these times negative one. That becomes a negative. This one becomes a positive. Now, negative 47x plus 50x is positive three x. And then 10x squared minus 10x squared gets canceled out. Bring down that 15. Come on down. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | Now, negative 47x plus 50x is positive three x. And then 10x squared minus 10x squared gets canceled out. Bring down that 15. Come on down. I used to watch a lot of Prices Right growing up. Never quite made it to the show. All right, x goes into three x how many times? |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | Come on down. I used to watch a lot of Prices Right growing up. Never quite made it to the show. All right, x goes into three x how many times? It goes three times. Three times negative five is negative 15. Three times x is three x. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | All right, x goes into three x how many times? It goes three times. Three times negative five is negative 15. Three times x is three x. We wanna subtract the orange from the teal. And so this becomes a negative. This becomes a positive. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | Three times x is three x. We wanna subtract the orange from the teal. And so this becomes a negative. This becomes a positive. 15, or negative 15 plus 15 is zero. And three x minus three x is zero. So you're just left with zero. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | This becomes a positive. 15, or negative 15 plus 15 is zero. And three x minus three x is zero. So you're just left with zero. So no remainder. So this whole thing you could re-express or simplify as two x squared plus 10x plus three. And once again, if this was on Khan Academy, there would be a little bit of an input box that looks something like this and you would have to type this in. |
Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3 | So you're just left with zero. So no remainder. So this whole thing you could re-express or simplify as two x squared plus 10x plus three. And once again, if this was on Khan Academy, there would be a little bit of an input box that looks something like this and you would have to type this in. Now, if you wanted these to be exactly the same expression, you would also need to constrain the domain. You would say, okay, four x does not equal positive five. And the reason why you have to constrain that is the whole reason why we can even divide by x minus five is we're assuming that x minus five is not equal to zero. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Welcome to this presentation on logarithm properties. Now this is going to be a very hands-on presentation. If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties. But what I'm going to do is I'm going to show you the properties and then show you how they can be used. It's going to be a little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a, oh, that's not the right, let's see, I want to change, there you go. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | But what I'm going to do is I'm going to show you the properties and then show you how they can be used. It's going to be a little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a, oh, that's not the right, let's see, I want to change, there you go. Let's say I say that a, let me start over, a to the b is equal to c. a to the b to the power is equal to c. So another way to write this exact same relationship, instead of writing an exponent, is to write it as a logarithm. So we could say that the logarithm base a of c is equal to b. So these are essentially saying the same thing, they just have different kind of results. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So if I say that a, oh, that's not the right, let's see, I want to change, there you go. Let's say I say that a, let me start over, a to the b is equal to c. a to the b to the power is equal to c. So another way to write this exact same relationship, instead of writing an exponent, is to write it as a logarithm. So we could say that the logarithm base a of c is equal to b. So these are essentially saying the same thing, they just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is. So the exact same relationship, just dated in a different way. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So these are essentially saying the same thing, they just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is. So the exact same relationship, just dated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm, let me do a more cheerful color. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So the exact same relationship, just dated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm, let me do a more cheerful color. The logarithm, let's say, of any base, so let's just call the base, let's say b for base, logarithm base b of a plus logarithm base b of c. And this only works if we have the same basis, so that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, I don't know, examples. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So the first is that the logarithm, let me do a more cheerful color. The logarithm, let's say, of any base, so let's just call the base, let's say b for base, logarithm base b of a plus logarithm base b of c. And this only works if we have the same basis, so that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, I don't know, examples. So this is saying that, I'll switch to another color. Let's make mauve my mauve. I don't know, I never know how to say that properly. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Or let's just even try it out with some, I don't know, examples. So this is saying that, I'll switch to another color. Let's make mauve my mauve. I don't know, I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32. So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | I don't know, I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32. So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Let's see if that's true just trying out this number. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Let's see if that's true just trying out this number. And this really isn't a proof, but it'll give you a little bit of an intuition, I think, for what's going on around here. So we just used our property, this little property that I presented to you, and let's see if it works out. So log base 2 of 8. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Let's see if that's true just trying out this number. And this really isn't a proof, but it'll give you a little bit of an intuition, I think, for what's going on around here. So we just used our property, this little property that I presented to you, and let's see if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? 2 to the third power is equal to 8. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? 2 to the third power is equal to 8. So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | 2 to the third power is equal to 8. So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see, 2 to the fourth power is 16, 2 to the fifth power is 32. So this right here is 5, right? And 2 to the what power is equal to 256? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | 2 to what power is equal to 32? Let's see, 2 to the fourth power is 16, 2 to the fifth power is 32. So this right here is 5, right? And 2 to the what power is equal to 256? Well, let's see. Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | And 2 to the what power is equal to 256? Well, let's see. Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you, or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you, or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent? I don't know. I don't know the names of things. And that equals 2 to the eighth. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | The additive exponent? I don't know. I don't know the names of things. And that equals 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 to the third times 2 to the fifth, essentially. And on this side, you have them added to each other. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | And that equals 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 to the third times 2 to the fifth, essentially. And on this side, you have them added to each other. And what makes logarithms interesting is, and why it's a little confusing at first, and you can watch the proofs if you really want a kind of a rigorous, not even my proofs aren't rigorous, but if you want kind of a better explanation of how this works. But this should hopefully give you an intuition for why this property holds, right? Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | And on this side, you have them added to each other. And what makes logarithms interesting is, and why it's a little confusing at first, and you can watch the proofs if you really want a kind of a rigorous, not even my proofs aren't rigorous, but if you want kind of a better explanation of how this works. But this should hopefully give you an intuition for why this property holds, right? Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos. So let me show you another log property that's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | If you don't believe me, watch the proof videos. So let me show you another log property that's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot, just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of 1 ninth minus log base 3 of 81. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot, just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of 1 ninth minus log base 3 of 81. So this property tells us that this is the same thing as, well, I'm ending up with a big number. Log base 3 of 1 ninth divided by 81. So that's the same thing as 1 ninth times 1 over 81. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Let's say log base 3 of 1 ninth minus log base 3 of 81. So this property tells us that this is the same thing as, well, I'm ending up with a big number. Log base 3 of 1 ninth divided by 81. So that's the same thing as 1 ninth times 1 over 81. I used two large numbers for my example. But we'll move forward. So let's see. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So that's the same thing as 1 ninth times 1 over 81. I used two large numbers for my example. But we'll move forward. So let's see. 9 times 8 is 720. Right? 9 times 8 is 720. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So let's see. 9 times 8 is 720. Right? 9 times 8 is 720. So this is 1 over 729. So this is log base 3 over 1 over 729. So 3 to what power is equal to 1 ninth? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | 9 times 8 is 720. So this is 1 over 729. So this is log base 3 over 1 over 729. So 3 to what power is equal to 1 ninth? Well, 3 squared is equal to 9, right? 3 squared is equal to 9. So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So 3 to what power is equal to 1 ninth? Well, 3 squared is equal to 9, right? 3 squared is equal to 9. So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? The negative just inverts it. So this is equal to negative 2. Right? |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? The negative just inverts it. So this is equal to negative 2. Right? And then minus 3 to what power is equal to 81? Let's see. 3 to the third power is 27. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Right? And then minus 3 to what power is equal to 81? Let's see. 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus 6 power is equal to 1 over 729. So my question is, 3 to the minus 6 power, is that equal to 1 over 729? Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus 6 power is equal to 1 over 729. So my question is, 3 to the minus 6 power, is that equal to 1 over 729? Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here, but let's see. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here, but let's see. 3 to the third power. This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. |
Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3 | Because, well, we could look here, but let's see. 3 to the third power. This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. You can confirm it with a calculator if you don't believe me. Anyway, that's all the time I have in this video. In the next video, I'll introduce you to the last two logarithm properties. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | We're told an electronic circuit has two resistors with resistances R1 and R2 connected in parallel. The circuit's total resistance, R sub T or RT, is given by this formula. Suppose we increase the value of R1 while keeping R2 constant. What does the value of R sub T increase, decrease, or stay the same? So pause this video and see if you can answer this question. All right, now let's work through this together. And some of you might be familiar with the idea of an electronic circuit and resistors and what they represent, but you really don't need to understand that in order to understand what's going on in this expression. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | What does the value of R sub T increase, decrease, or stay the same? So pause this video and see if you can answer this question. All right, now let's work through this together. And some of you might be familiar with the idea of an electronic circuit and resistors and what they represent, but you really don't need to understand that in order to understand what's going on in this expression. There's some quantity, R sub T, that's equal to one over, and then in the denominator, we have one over R1 plus one over R2. So if we increase the value of R1 while keeping R2 constant, what happens? So this is going to increase, and R2 is going to be constant. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | And some of you might be familiar with the idea of an electronic circuit and resistors and what they represent, but you really don't need to understand that in order to understand what's going on in this expression. There's some quantity, R sub T, that's equal to one over, and then in the denominator, we have one over R1 plus one over R2. So if we increase the value of R1 while keeping R2 constant, what happens? So this is going to increase, and R2 is going to be constant. So one way to think about it, we have two variables here, especially in this denominator, but really in this entire expression. But if R2 is going to be constant, we really just have to focus our analysis on R1. If R2 is constant, that means it's just a number. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | So this is going to increase, and R2 is going to be constant. So one way to think about it, we have two variables here, especially in this denominator, but really in this entire expression. But if R2 is going to be constant, we really just have to focus our analysis on R1. If R2 is constant, that means it's just a number. It could be two, it could be five, it could be pi, whatever, but that is not going to change as we increase the value of R1. So let's think about what's happening here. If R sub one increases, then what does that do to one over R1? |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | If R2 is constant, that means it's just a number. It could be two, it could be five, it could be pi, whatever, but that is not going to change as we increase the value of R1. So let's think about what's happening here. If R sub one increases, then what does that do to one over R1? Well, if you increase the denominator, then you are going to decrease the reciprocal of that. So that means that this whole thing right over here is going to decrease. Now, if one over R1 is decreasing, if one over R1 is decreasing, what's going to happen to one over R1 plus one over R2? |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | If R sub one increases, then what does that do to one over R1? Well, if you increase the denominator, then you are going to decrease the reciprocal of that. So that means that this whole thing right over here is going to decrease. Now, if one over R1 is decreasing, if one over R1 is decreasing, what's going to happen to one over R1 plus one over R2? Will this entire expression increase or decrease? Well, this part is staying constant. R2 is constant, so one over R2 is constant. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | Now, if one over R1 is decreasing, if one over R1 is decreasing, what's going to happen to one over R1 plus one over R2? Will this entire expression increase or decrease? Well, this part is staying constant. R2 is constant, so one over R2 is constant. Just imagine R2 could be two or three, so this would just be 1 1⁄2 or 1 3rd or whatever it is, while over here, this part of the expression is going down. So if you're taking the sum of two things, one part's going down, the other part's constant, then that means this whole thing is going to be going down. So the entire denominator of this entire thing is going down. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | R2 is constant, so one over R2 is constant. Just imagine R2 could be two or three, so this would just be 1 1⁄2 or 1 3rd or whatever it is, while over here, this part of the expression is going down. So if you're taking the sum of two things, one part's going down, the other part's constant, then that means this whole thing is going to be going down. So the entire denominator of this entire thing is going down. Now, if the entire denominator is going down, if one over R1 plus one over R2, if this whole thing is going down, what's going to happen to the reciprocal of that? One over, one over R1 plus one over R2. Well, if something is going down, the reciprocal of that is going to go up. |
Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3 | So the entire denominator of this entire thing is going down. Now, if the entire denominator is going down, if one over R1 plus one over R2, if this whole thing is going down, what's going to happen to the reciprocal of that? One over, one over R1 plus one over R2. Well, if something is going down, the reciprocal of that is going to go up. If you get smaller and smaller denominators, one over that is going to be a larger and larger value. So the value of RT increases if R1 increases and R2 is constant. And for those of you who know about resistance, which is really how well a current can flow through a circuit that will also make intuitive sense, but you don't need to understand resistance to analyze this mathematically. |
Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3 | So immediately you might say, well, this is either going to be a sine function or cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum point, it hits a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4 divided by 2 is negative 2. |