1971 AHSME Problems/Problem 21

Problem

If $\log_2(\log_3(\log_4 x))=\log_3(\log_4(\log_2 y))=\log_4(\log_2(\log_3 z))=0$ , then the sum $x+y+z$ is equal to

$\textbf{(A) }50\qquad \textbf{(B) }58\qquad \textbf{(C) }89\qquad \textbf{(D) }111\qquad \textbf{(E) }1296$

If $\log_{x}{n}=0,$ then $n = 1.$ So, we can rewrite this equation: $(\log_3(\log_4 x))=(\log_4(\log_2 y))=(\log_2(\log_3 z))=1$

Solve individually for each variable. $x = 4^3 = 64$ $y = 2^4 = 16$ $z = 3^2 = 9$ Therefore, $x + y + z = 89.$

The answer is $\boxed{\textbf{(C) }89}.$

-edited by coolmath34

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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