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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>
  
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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 $a=\log_8 225$ and $b=\log_2 15$, then

$\text{(A) } a=b/2\quad \text{(B) } a=2b/3\quad \text{(C) } a=b\quad \text{(D) } b=a/2\quad \text{(E) } a=3b/2$

Solution

The solutions imply that finding the ratio $\frac{a}{b}$ will solve the problem. We compute $\frac{a}{b}$, use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base:

$\frac{\log_8 225}{\log_2 15}$

$\frac{\frac{\ln 225}{\ln 8}}{\frac{\ln 15}{\ln 2}}$

$\frac{\ln 225 \ln 2}{\ln 15 \ln 8}$

$\frac{\ln 225}{\ln 15} \cdot \frac{\ln 2}{\ln 8}$

$\ln_{15} 225 \cdot \ln_8 2$

Since $15^2 = 225$, the first logarithm is $2$. Since $8^{\frac{1}{3}} = 2$, the second logarithm is $\frac{1}{3}$.

Thus, we have $\frac{a}{b} = \frac{2}{3}$, or $a = \frac{2}{3}b$, which is option $\fbox{B}$.

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
All AHSME Problems and Solutions

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