1960 AHSME Problems/Problem 24

Revision as of 20:55, 11 May 2018 by Rockmanex3 (talk | contribs) (Solution for Problem 24)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
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 36 37 38 39 40
All AHSME Problems and Solutions
Invalid username
Login to AoPS