HMMT February, 2020

Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:
Let $a = 256$ . Find the unique real number $x > a^{2}$ such that
$\log_{a} \log_{a} \log_{a} x = \log_{a^{2}} \log_{a^{2}} \log_{a^{2}} x$

Solution

Solution:
Let $y = \log_{a} x$ so $\log_{a} \log_{a} y = \log_{a^{2}} \log_{a^{2}} \frac{1}{2} y$ . Setting $z = \log_{a} y$ , we find $\log_{a} z = \log_{a^{2}}\left(\frac{1}{2} z - \frac{1}{16}\right)$ , or $z^{2} - \frac{1}{2} z + \frac{1}{16} = 0$ . Thus, we have $z = \frac{1}{4}$ , so we can backsolve to get $y = 4$ and $x = 2^{32}$ .

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