1962 AHSME Problems/Problem 17: Difference between revisions

Revision as of 09:31, 17 April 2014

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

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

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)

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

@@ Line 5: / Line 5: @@
 ==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>.
+<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 \log_{15} 225 \log_8 2</cmath>
+<cmath>a = \boxed{\frac{2b}3 \textbf{ (B)}}</cmath>