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

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<cmath>a = b \log_{15} 225 \log_8 2</cmath>
 
<cmath>a = b \log_{15} 225 \log_8 2</cmath>
 
<cmath>a = \boxed{\frac{2b}3 \textbf{ (B)}}</cmath>
 
<cmath>a = \boxed{\frac{2b}3 \textbf{ (B)}}</cmath>
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 16|num-a=18}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:17, 3 October 2014

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 $b = \frac{\log 15}{\log 2}$. \[\frac{a}b = \frac{\log 225 \log 2}{\log 8 \log 15}\]

\[a = b \dot \frac{\log 225 \log 15} \dot \frac{\log 2 \log 8}\] (Error compiling LaTeX. Unknown error_msg)

\[a = b \log_{15} 225 \log_8 2\] \[a = \boxed{\frac{2b}3 \textbf{ (B)}}\]

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
All AHSME Problems and Solutions

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