HMMT February

Algebra Difficulty 4.9 AIME Find the answer

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

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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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