Difference between revisions of "1962 AHSME Problems/Problem 17"

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==Solution==
 
==Solution==
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Using the change-of-base rule: <math>a = \frac{\log 225}{\log 8}</math> and <math>b = \frac{\log 15}{\log 2}</math>.
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<cmath>\frac{a}{b} = \frac{\log 225 \log 2}{\log 8 \log 15}</cmath>
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<cmath>a = b \cdot \frac{\log 225}{\log 15} \cdot \frac{\log 2}{\log 8}</cmath>
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<cmath>a = b \log_{15} 225 \log_8 2</cmath>
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<cmath>a = \boxed{\frac{2b}3 \textbf{ (B)}}</cmath>
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==See Also==
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{{AHSME 40p box|year=1962|num-b=16|num-a=18}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:48, 31 May 2018

Problem

If $a = \log_8 225$ and $b = \log_2 15$, then $a$, in terms of $b$, is:

$\textbf{(A)}\ \frac{b}{2}\qquad\textbf{(B)}\ \frac{2b}{3}\qquad\textbf{(C)}\ b\qquad\textbf{(D)}\ \frac{3b}{2}\qquad\textbf{(E)}\ 2b$

Solution

Using the change-of-base rule: $a = \frac{\log 225}{\log 8}$ and $b = \frac{\log 15}{\log 2}$. \[\frac{a}{b} = \frac{\log 225 \log 2}{\log 8 \log 15}\] \[a = b \cdot \frac{\log 225}{\log 15} \cdot \frac{\log 2}{\log 8}\] \[a = b \log_{15} 225 \log_8 2\] \[a = \boxed{\frac{2b}3 \textbf{ (B)}}\]


See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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
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