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

(Created page with "==Problem 5== If <math>\log_6 x=2.5</math>, the value of <math>x</math> is: <math>\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 36 \qquad \textbf{(C)}\ 36\sqrt{6} \qquad \textbf{(D...")
 
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== Solution ==
 
== 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>
 
<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>
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== Solution 2 ==
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<math>\log_6 x=\frac{5}{2}\implies 6^{\frac{5}{2}}=x \implies \sqrt{6^5}=x \implies 36\sqrt{6} = x</math>
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==See Also==
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{{AHSME 50p box|year=1953|num-b=4|num-a=6}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 20:21, 1 April 2017

Problem 5

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

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

Solution

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

Solution 2

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


See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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