AMC12 2014 12A, Problem 18

The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers.  What is $m+n$?

Solution: For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that: $\log_b a$ is defined if and only if $a>0.$ For $0<b<1,$ we conclude that: $\log_b a<c$ if and only if $a>b^c.$ $\log_b a>c$ if and only if $0<a<b^c.$ For $b>1,$ we conclude that: $\log_b a<c$ if and only if $0<a<b^c.$ $\log_b a>c$ if and only if $a>b^c.$ Therefore, we have \begin{align} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align} The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\fbox{271}.$ Remark This problem is quite similar to 2004 AMC 12A Problem 16.

Answer: 271