Difference between revisions of "1961 AHSME Problems/Problem 6"
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}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 20:01, 17 May 2018
Problem
When simplified, 
becomes:
Solution
First, note that 
. That means the expression can be rewritten as
![\[\log{8} \div \log{8^{-1}}\]](http://latex.artofproblemsolving.com/c/2/2/c226ad7bb6269a4dfb1ddb3643fe92d196a17cda.png)
![\[\log{8} \cdot \frac{1}{\log{8^{-1}}}\]](http://latex.artofproblemsolving.com/1/4/9/1499798c40200f63c1301698f68cc792d9151b9b.png)
![\[\log{8} \cdot \frac{1}{-1 \cdot \log{8}}\]](http://latex.artofproblemsolving.com/a/3/c/a3c102a9fee21d1123fa30270bb00f36ddc699dc.png)
This simplifies to 
, which is answer choice 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
