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

m (→‎Solution)
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==Solution==
 
==Solution==
 
Using the change-of-base rule: <math>a = \frac{\log 225}{\log 8}</math> and <math>b = \frac{\log 15}{\log 2}</math>.
 
Using the change-of-base rule: <math>a = \frac{\log 225}{\log 8}</math> and <math>b = \frac{\log 15}{\log 2}</math>.
−
<cmath>\frac{a}b = \frac{\log 225 \log 2}{\log 8 \log 15}</cmath>
+
<cmath>\frac{a}{b} = \frac{\log 225 \log 2}{\log 8 \log 15}</cmath>
−
<cmath>a = b \dot \frac{\log 225 \log 15} \dot \frac{\log 2 \log 8}</cmath>
+
<cmath>a = b \cdot \frac{\log 225}{\log 15} \cdot \frac{\log 2}{\log 8}</cmath>
 
<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>
 +
  
 
==See Also==
 
==See Also==
−
{{AHSME 40p box|year=1962|before=Problem 16|num-a=18}}
+
{{AHSME 40p box|year=1962|num-b=16|num-a=18}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{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 (Problems • Answer Key • Resources)
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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