Difference between revisions of "1961 AHSME Problems/Problem 6"
(Created page with 'When simplified, <math>\log 8 \div \log {\frac{1}{8}}</math> becomes:') |
Rockmanex3 (talk | contribs) (Solution to Problem 6) |
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− | When simplified, <math>\log 8 \div \log {\frac{1}{8}}</math> becomes: | + | == Problem == |
+ | |||
+ | When simplified, <math>\log{8} \div \log{\frac{1}{8}}</math> becomes: | ||
+ | |||
+ | <math>\textbf{(A)}\ 6\log{2} \qquad | ||
+ | \textbf{(B)}\ \log{2} \qquad | ||
+ | \textbf{(C)}\ 1 \qquad | ||
+ | \textbf{(D)}\ 0\qquad | ||
+ | \textbf{(E)}\ -1 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | First, note that <math>\frac{1}{8} = 8^{-1}</math>. That means the expression can be rewritten as | ||
+ | <cmath>\log{8} \div \log{8^{-1}}</cmath> | ||
+ | <cmath>\log{8} \cdot \frac{1}{\log{8^{-1}}}</cmath> | ||
+ | <cmath>\log{8} \cdot \frac{1}{-1 \cdot \log{8}}</cmath> | ||
+ | This simplifies to <math>-1</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1961|num-b=5|num-a=7}} | ||
+ | |||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 19:01, 17 May 2018
Problem
When simplified, becomes:
Solution
First, note that . That means the expression can be rewritten as This simplifies to , which is answer choice .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AHSME Problems and Solutions |
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