Difference between revisions of "1970 AHSME Problems/Problem 8"
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<math>\ln_{15} 225 \cdot \ln_8 2</math> | <math>\ln_{15} 225 \cdot \ln_8 2</math> | ||
− | Since <math>15^2 = 225</math>, the first logarithm is <math>2</math>. Since <math>8^{\frac{1}{3} = 2</math>, the second logarithm is <math>\frac{1}{3}</math>. | + | Since <math>15^2 = 225</math>, the first logarithm is <math>2</math>. Since <math>8^{\frac{1}{3}} = 2</math>, the second logarithm is <math>\frac{1}{3}</math>. |
Thus, we have <math>\frac{a}{b} = \frac{2}{3}</math>, or <math>a = \frac{2}{3}b</math>, which is option <math>\fbox{B}</math>. | Thus, we have <math>\frac{a}{b} = \frac{2}{3}</math>, or <math>a = \frac{2}{3}b</math>, which is option <math>\fbox{B}</math>. |
Latest revision as of 20:02, 13 July 2019
Problem
If and , then
Solution
The solutions imply that finding the ratio will solve the problem. We compute , use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base:
Since , the first logarithm is . Since , the second logarithm is .
Thus, we have , or , which is option .
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.