Maths Olympiad Prep

Library / HMMT February /481 of 1302

HMMT February

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. Fractions can be typed as 3/2, and spacing doesn't matter.

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}.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.