2014 AMC 12A Problem 18

Question

The domain of the function

\tiny f(x)=\log_{\frac{1}{2}}\!\left(\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{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

Peel the logarithms from the outside in, requiring each argument to be positive

Remember that bases below

1

reverse inequalities; translate each successive positivity condition before moving to the next inner logarithm

Working from the outside,

f

is defined exactly when

\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\right)

\gt0,

which is equivalent to

\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\gt1.

Since the base

\tfrac14\lt1,

this means

0\lt\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\lt\tfrac14,

hence

1\lt\log_{\frac{1}{16}}x\lt16^{\frac{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.

Problem 18 in other years