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Algebra Difficulty 5.0 AIME Prove it United States

Problem:
Let $a$ and $b$ be real numbers greater than $1$ such that $a b = 100$ . The maximum possible value of $a^{\left(\log_{10} b\right)^2}$ can be written in the form $10^{x}$ for some real number $x$ . Find $x$ .

Solution

Solution:
Let $p = \log_{10} a$ , $q = \log_{10} b$ . Since $a, b > 1$ , $p$ and $q$ are positive. The condition $a b = 100$ translates to $p + q = 2$ . We wish to maximize
$x = \log_{10} a^{\left(\log_{10} b\right)^2} = \left(\log_{10} a\right)\left(\log_{10} b\right)^2 = p q^2$
By AM-GM,
$\frac{27}{4} p q^2 \leq \left(p + \frac{q}{2} + \frac{q}{2}\right)^3 = 8$
Hence $p q^2 \leq \frac{32}{27}$ with equality when $p = \frac{2}{3}$ , $q = \frac{4}{3}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.