[Algebra]
Problem: The populations of Alphaville and Betaville were equal at the end of 1995. The population of Alphaville decreased by $2.9 \%$ during 1996, then increased by $8.9 \%$ during 1997 , and then increased by $6.9 \%$ during 1998 . The population of Betaville increased by $r \%$ in each of the three years. If the populations of the towns are equal at the end of 1998, determine the value of $r$ correct to one decimal place.

Solution: If $P$ is the original population of Alphaville and Betaville,

$$
\begin{aligned}
P(.971)(1.089)(1.069) & =P\left(1+\frac{r}{100}\right)^{3} \\
1.1303 & =\left(1+\frac{r}{100}\right)^{3}
\end{aligned}
$$

From here,

Possibility 1

$$
\begin{aligned}
1+\frac{r}{100} & =(1.1303)^{\frac{1}{3}} \\
1+\frac{r}{100} & =1.0416 \\
r & \doteq 4.2 \%
\end{aligned}
$$

Or, Possibility 2

$$
\begin{aligned}
3 \log \left(1+\frac{r}{100}\right) & =\log 1.1303 \\
\log \left(1+\frac{r}{100}\right) & =.01773 \\
1+\frac{r}{100} & =1.0416 \\
r & \doteq 4.2 \%
\end{aligned}
$$

Final answer: 4.2 %