Difference between revisions of "1953 AHSME Problems/Problem 5"

Revision as of 19:21, 1 April 2017

If $\log_6 x=2.5$ , the value of $x$ is:

$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 36 \qquad \textbf{(C)}\ 36\sqrt{6} \qquad \textbf{(D)}\ 0.5 \qquad \textbf{(E)}\ \text{none of these}$

$\log_6 x=\frac{5}{2}\implies 2\log_6 x=5\implies \log_6 x^2=5\implies x^2=6^5\implies x=36\sqrt{6}$ , $\fbox{C}$

$\log_6 x=\frac{5}{2}\implies 6^{\frac{5}{2}}=x \implies \sqrt{6^5}=x \implies 36\sqrt{6} = x$

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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 == Solution ==
 <math>\log_6 x=\frac{5}{2}\implies 2\log_6 x=5\implies \log_6 x^2=5\implies x^2=6^5\implies x=36\sqrt{6}</math>, <math>\fbox{C}</math>
+== Solution 2 ==
+<math>\log_6 x=\frac{5}{2}\implies 6^{\frac{5}{2}}=x \implies \sqrt{6^5}=x \implies 36\sqrt{6} = x</math>
+==See Also==
+{{AHSME 50p box|year=1953|num-b=4|num-a=6}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}