Question

The domain of the function $f(x)=\log_{1/2}\!\left(\log_4\!\left(\log_{1/4}\!\left(\log_{16}\!\left(\log_{1/16}x\right)\right)\right)\right)$ is an interval of length $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$19$

$31$

$271$

$319$

$511$

Solution:

Working from the outside, $f$ is defined exactly when $\log_4\!\left(\log_{1/4}\!\left(\log_{16}\!\left(\log_{1/16}x\right)\right)\right)\gt0,$ which is equivalent to $\log_{1/4}\!\left(\log_{16}\!\left(\log_{1/16}x\right)\right)\gt1.$

Since the base $\tfrac14\lt1,$ this means $0\lt\log_{16}\!\left(\log_{1/16}x\right)\lt\tfrac14,$ hence $1\lt\log_{1/16}x\lt16^{1/4}=2.$

As $\tfrac{1}{16}\lt1,$ this reverses to $\left(\tfrac{1}{16}\right)^2\lt x\lt\left(\tfrac{1}{16}\right)^1,$ i.e. $\tfrac{1}{256}\lt x\lt\tfrac{1}{16}.$ The length is $\tfrac{1}{16}-\tfrac{1}{256}=\tfrac{15}{256},$ so $m+n=15+256=271.$

Thus, the correct answer is C.

2014 AMC 12A Problem 18

Solution:

Problem 18 in Other Years