We assume a relationship of the form
$\begin{align*} \log_{10} G = m \log_{10} \omega + b\end{align*}$
and attempt to calculate $m$ . Raising $10$ to the power of both sides, we see that $G = 10^b\omega^m$ , so finding $m$ will allow us to distinguish between answers (B), (C), (D), and (E). From the two points $(3 \cdot 10^5, 10^2)$ and $(2 \cdot 10^5, 2 \cdot 10^2)$ , we find that
$\begin{align*}m = - \frac{\log_{10} \left(2\right)}{\log_{10} \left(3/2\right)} = - \frac{\log_{2} \left(2\right)}{\log_{2} \...$
Since $(3/2)^2 = 9/4 \approx 2$ , $1/2$ provides a better approximation to $\log_2 \left(3/2\right)$ than $1$ . Therefore,
$\begin{align*}m \approx -2\end{align*}$
and answer (E) is correct.

Solution to 1986 Problem 39