Difference between revisions of "1960 AHSME Problems/Problem 24"

(Solution for Problem 24)
 
(Fix the format)
 
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Rewrite the equation to <math>(2x)^x = 216</math>.  Since <math>2x</math> is the base of the logarithm, we only need to check x-values that are positive.
 
Rewrite the equation to <math>(2x)^x = 216</math>.  Since <math>2x</math> is the base of the logarithm, we only need to check x-values that are positive.
  
With trial and error, <math>3</math> is a solution because <math>(2 \cdot 3)^3 = 6^3 = 216.  Since </math>(2x)^x<math> gets larger as x gets larger, </math>3<math> is the only solution, so the answer is </math>\boxed{\textbf{(A)}}$.
+
With trial and error, <math>3</math> is a solution because <math>(2 \cdot 3)^3 = 6^3 = 216</math>.  Since <math>(2x)^x</math> gets larger as x gets larger, <math>3</math> is the only solution, so the answer is <math>\boxed{\textbf{(A)}}</math>.
 
 
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=23|num-a=25}}
 
{{AHSME 40p box|year=1960|num-b=23|num-a=25}}

Latest revision as of 20:55, 11 May 2018

Problem

If $\log_{2x}216 = x$, where $x$ is real, then $x$ is:

$\textbf{(A)}\ \text{A non-square, non-cube integer}\qquad$ $\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad$ $\textbf{(C)}\ \text{An irrational number}\qquad$ $\textbf{(D)}\ \text{A perfect square}\qquad$ $\textbf{(E)}\ \text{A perfect cube}$

Solution

Rewrite the equation to $(2x)^x = 216$. Since $2x$ is the base of the logarithm, we only need to check x-values that are positive.

With trial and error, $3$ is a solution because $(2 \cdot 3)^3 = 6^3 = 216$. Since $(2x)^x$ gets larger as x gets larger, $3$ is the only solution, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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