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

Latest revision as of 08:58, 5 August 2024

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 19: / Line 19: @@
 Therefore, <math>x + y + z = 89.</math>
-The answer is <math>\textbf{(C)}.</math>
+The answer is <math>\boxed{\textbf{(C) }89}.</math>
 -edited by coolmath34
+== See Also ==
+{{AHSME 35p box|year=1971|num-b=20|num-a=22}}
+{{MAA Notice}}