# Difference between revisions of "1984 AIME Problems/Problem 5"

## Problem

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$.

## Solution

Use the change of base formula to see that $\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5$; combine denominators to find that $\frac{\log a^3b}{3\log 2} = 5$. Doing the same thing with the second equation yields that $\frac{\log b^3a}{3\log 2} = 7$. This means that $\log a^3b = 15\log 2 \Longrightarrow a^3b = 2^{15}$ and that $\log ab^3 = 21\log 2 \Longrightarrow ab^3 = 2^{21}$. If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$, so taking the fourth root of that, $ab = 2^9 = \boxed{512}$.