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

(Created page with "==Problem== If <math>a = \log_8 225</math> and <math>b = \log_2 15</math>, then <math>a</math>, in terms of <math>b</math>, is: <math> \textbf{(A)}\ \frac{b}{2}\qquad\textbf{(B...")
 
(Fixing some stuff)
 
(3 intermediate revisions by 3 users not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
"Unsolved"
+
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>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 = \boxed{\frac{2b}3 \textbf{ (B)}}</cmath>
 +
 
 +
 
 +
==See Also==
 +
{{AHSME 40p box|year=1962|num-b=16|num-a=18}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:48, 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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png