Difference between revisions of "1961 AHSME Problems/Problem 6"

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When simplified, <math>\log 8 \div \log {\frac{1}{8}}</math> becomes:
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== Problem ==
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When simplified, <math>\log{8} \div \log{\frac{1}{8}}</math> becomes:
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<math>\textbf{(A)}\ 6\log{2} \qquad
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\textbf{(B)}\ \log{2} \qquad
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\textbf{(C)}\ 1 \qquad
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\textbf{(D)}\ 0\qquad
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\textbf{(E)}\ -1 </math> 
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==Solution==
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First, note that <math>\frac{1}{8} = 8^{-1}</math>.  That means the expression can be rewritten as
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<cmath>\log{8} \div \log{8^{-1}}</cmath>
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<cmath>\log{8} \cdot \frac{1}{\log{8^{-1}}}</cmath>
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<cmath>\log{8} \cdot \frac{1}{-1 \cdot \log{8}}</cmath>
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This simplifies to <math>-1</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=5|num-a=7}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:01, 17 May 2018

Problem

When simplified, $\log{8} \div \log{\frac{1}{8}}$ becomes:

$\textbf{(A)}\ 6\log{2} \qquad \textbf{(B)}\ \log{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0\qquad \textbf{(E)}\ -1$

Solution

First, note that $\frac{1}{8} = 8^{-1}$. That means the expression can be rewritten as \[\log{8} \div \log{8^{-1}}\] \[\log{8} \cdot \frac{1}{\log{8^{-1}}}\] \[\log{8} \cdot \frac{1}{-1 \cdot \log{8}}\] This simplifies to $-1$, which is answer choice $\boxed{\textbf{(E)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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


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