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Difference between revisions of "1971 AHSME Problems/Problem 21"

(Created page with "== Problem == If <math>\log_2(\log_3(\log_4 x))=\log_3(\log_4(\log_2 y))=\log_4(\log_2(\log_3 z))=0</math>, then the sum <math>x+y+z</math> is equal to <math>\textbf{(A) }50...")
(No difference)

Revision as of 20:10, 28 January 2021

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

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

Solution

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 $\textbf{(C)}.$

-edited by coolmath34

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