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Algebra Difficulty 5.2 AIME, harder Find the answer

Suppose a real number x>1x>1 satisfies log⁡2(log⁡4x)+log⁡4(log⁡16x)+log⁡16(log⁡2x)=0\log _{2}\left(\log _{4} x\right)+\log _{4}\left(\log _{16} x\right)+\log _{16}\left(\log _{2} x\right)=0. Compute log⁡2(log⁡16x)+log⁡16(log⁡4x)+log⁡4(log⁡2x)\log _{2}\left(\log _{16} x\right)+\log _{16}\left(\log _{4} x\right)+\log _{4}\left(\log _{2} x\right).

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let AA and BB be these sums, respectively. Then B−A=log⁡2(log⁡16xlog⁡4x)+log⁡4(log⁡2xlog⁡16x)+log⁡16(log⁡4xlog⁡2x)=log⁡2(log⁡164)+log⁡4(log⁡216)+log⁡16(log⁡42)=log⁡2(12)+log⁡44+log⁡16(12)=(−1)+1+(−14)=−14B-A =\log _{2}\left(\frac{\log _{16} x}{\log _{4} x}\right)+\log _{4}\left(\frac{\log _{2} x}{\log _{16} x}\right)+\log _{16}\left(\frac{\log _{4} x}{\log _{2} x}\right) =\log _{2}\left(\log _{16} 4\right)+\log _{4}\left(\log _{2} 16\right)+\log _{16}\left(\log _{4} 2\right) =\log _{2}\left(\frac{1}{2}\right)+\log _{4} 4+\log _{16}\left(\frac{1}{2}\right) =(-1)+1+\left(-\frac{1}{4}\right) =-\frac{1}{4}. Since A=0A=0, we have the answer B=−14B=-\frac{1}{4}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.