[Algebra]
Problem: A chef aboard a luxury liner wants to cook a goose. The time $t$ in hours to cook a goose at $180^{\circ} \mathrm{C}$ depends on the mass of the goose $m$ in kilograms according to the formula

$$
t=a m^{b}
$$

where $a$ and $b$ are constants. The table below gives the times observed to cook a goose at $180^{\circ} \mathrm{C}$.

| Mass, $m(\mathrm{~kg})$ | Time, $t(\mathrm{~h})$ |
| :---: | :---: |
| 3.00 | 2.75 |
| 6.00 | 3.75 |
Using the data in the table, determine both $a$ and $b$ to two decimal places.

Solution: From the table we have two pieces of information, so we substitute both of these into the given formula.

$$
\begin{aligned}
& 2.75=a(3.00)^{b} \\
& 3.75=a(6.00)^{b}
\end{aligned}
$$

We can now proceed in either of two ways to solve for $b$.

Method 1 to find $b$

Dividing the second equation by the first, we obtain

$$
\frac{3.75}{2.75}=\frac{a(6.00)^{b}}{a(3.00)^{b}}=\frac{(6.00)^{b}}{(3.00)^{b}}=\left(\frac{6.00}{3.00}\right)^{b}=2^{b}
$$

or

$$
2^{b} \approx 1.363636
$$

Taking logarithms of both sides,



$$
\begin{aligned}
\log \left(2^{b}\right) & \approx \log (1.363636) \\
b \log (2) & \approx \log (1.363636) \\
b & \approx \frac{\log (1.363636)}{\log (2)} \\
b & \approx 0.4475
\end{aligned}
$$

Method 2 to find $b$ 

Taking logarithms of both sides of the above equations, we obtain

$$
\begin{aligned}
\log (2.75) & =\log \left(a(3.00)^{b}\right) \\
& =\log (a)+\log \left((3.00)^{b}\right) \\
& =\log (a)+b \log (3.00)
\end{aligned}
$$

Similarly,

$$
\log (3.75)=\log (a)+b \log (6.00)
$$

Subtracting the first equation from the second, we obtain

$$
\begin{aligned}
\log (3.75)-\log (2.75) & =b(\log (6.00)-\log (3.00)) \\
b & =\frac{\log (3.75)-\log (2.75)}{\log (6.00)-\log (3.00)} \\
b & \approx 0.4475
\end{aligned}
$$

We now continue in the same way for both methods.

Substituting this value for $b$ back into the first equation above,

$$
\begin{aligned}
2.75 & \approx a(3.00)^{0.4475} \\
a & \approx \frac{2.75}{(3.00)^{0.4475}} \\
a & \approx 1.6820
\end{aligned}
$$

Therefore, to two decimal places, $a=1.68$ and $b=0.45$.

Final answer: $1.68,0.45$