INPUT ARTICLE: Article: This will tell us the hour of the day. The hours hand always points at the big numbers on the clock. When it's pointed between two numbers, the hour of the day is always the lower number. So if the hours hand is pointed between 5 and 6 on the clock, it's 5-something, because 5 is the lower number. For example, if the small, thick hand is pointed directly at the number 9, it's exactly 9 o'clock. Watch as the hour hand moves slowly closer to a big number, the minutes hand gets closer to the number 12. When the minutes hand hits 12, the next hour starts.

SUMMARY: Look at the number that the hours hand (the small, thick hand) is pointing at. Know that often, the hours hand will be pointed in between two numbers. Know that if the hours hand is pointed directly at a big number, it's exactly that-o'clock. Learn the movement of the hands.

INPUT ARTICLE: Article: This method covers problems in the form logb(x)logb(a){\displaystyle {
rac {\log _{b}(x)}{\log _{b}(a)}}}. However, it does not work for a few special cases: The log of a negative number is undefined for all bases (such as log(3){\displaystyle \log(-3)} or log4(5){\displaystyle \log _{4}(-5)}). Write "no solution." The log of zero is also undefined for all bases. If you see a term such as ln(0){\displaystyle \ln(0)}, write "no solution." The log of one in any base (log(1){\displaystyle \log(1)}) always equals zero, since x0=1{\displaystyle x^{0}=1} for all values of x. Replace that logarithm with 1 instead of using the method below. If the two logarithms have different bases, such as log3(x)log4(a){\displaystyle {
rac {log_{3}(x)}{log_{4}(a)}}}, and you cannot simplify either one into an integer, the problem is not feasible to solve by hand. Assuming you did not find any of the exceptions above, you can now simplify the problem into one logarithm. To do this, use the formula logb(x)logb(a)=loga(x){\displaystyle {
rac {\log _{b}(x)}{\log _{b}(a)}}=\log _{a}(x)}. Example 1: Solve the problem log16log2{\displaystyle {
rac {\log {16}}{\log {2}}}}.Start by converting this into one logarithm using the formula above: log16log2=log2(16){\displaystyle {
rac {\log {16}}{\log {2}}}=\log _{2}(16)}. This formula is the "change of base" formula, derived from basic logarithmic properties. Remember, to solve loga(x){\displaystyle \log _{a}(x)}, think "a?=x{\displaystyle a^{?}=x}" or "What exponent can I raise a by to get x?" It's not always feasible to solve this without a calculator, but if you're lucky, you'll end up with an easily simplified logarithm. Example 1 (cont.): Rewrite log2(16){\displaystyle \log _{2}(16)} as 2?=16{\displaystyle 2^{?}=16}. The value of "?" is the answer to the problem. You may need to find it by trial and error:22=22=4{\displaystyle 2^{2}=2*2=4}23=42=8{\displaystyle 2^{3}=4*2=8}24=82=16{\displaystyle 2^{4}=8*2=16}16 is what you were looking for, so log2(16){\displaystyle \log _{2}(16)} = 4. Some logarithms are very difficult to solve by hand. You'll need a calculator if you need the answer for a practical purpose. If you're solving problems in math class, your teacher most likely expects you to leave the answer as a logarithm. Here's another example using this method on a more difficult problem: Example 2: What is log3(58)log3(7){\displaystyle {
rac {\log _{3}(58)}{\log _{3}(7)}}}? Convert this into one logarithm: log3(58)log3(7)=log7(58){\displaystyle {
rac {\log _{3}(58)}{\log _{3}(7)}}=\log _{7}(58)}. (Notice that the 3 in each initial log disappears; this is true for any base.) Rewrite as 7?=58{\displaystyle 7^{?}=58} and test possible values of?:72=77=49{\displaystyle 7^{2}=7*7=49}73=497=343{\displaystyle 7^{3}=49*7=343}Since 58 falls between these two numbers, log7(58){\displaystyle \log _{7}(58)} has no integer answer. Leave your answer as log7(58){\displaystyle \log _{7}(58)}.

SUMMARY:
Check for negative numbers and ones. Convert the expression into one logarithm. Calculate by hand if possible. Leave the answer in logarithm form if you cannot simplify it.