AIME 1984 P5

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Problem Statement

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

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Solution

$$\log_8a+\log_4b^2=5 \text{(1)}$$ $$\log_8b+\log_4a^2=7 \text{(2)}$$ $$\text{(1): } \dfrac{\log_2a}{\log_2(8)} + \dfrac{\log_2b^2}{log_2(4)} = 5$$ $$\dfrac{\log_2a}{3} + \dfrac{\log_2b^2}{2} = 5$$ $$2\log_2a + 3\log_2b^2 = 30 \text{(3)}$$ $$\text{(2): } \dfrac{\log_2b}{\log_2(8)} + \dfrac{\log_2a^2}{log_2(4)} = 7$$ $$\dfrac{\log_2b}{3} + \dfrac{\log_2a^2}{2} = 7$$ $$2\log_2b + 3\log_2a^2 = 42 \text{(4)}$$ $$\text{(3) + (4): } 2(\log_2(a) + \log_2(b)) + 3(\log_2(a^2) + \log_2(b^2)) = 72$$ $$ 2\log_2ab + 3\log_2(ab)^2 = 72$$ $$ 2\log_2ab + 6\log_2(ab) = 72$$ $$ 8\log_2ab = 72$$ $$ \log_2ab = 9 \implies 2^9 = ab$$ $$\therefore ab = 512$$