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1984 AIME Problems/Problem 5

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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 = 512.

See also

1984 AIME (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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