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...") |
Rockmanex3 (talk | contribs) (Fixing some stuff) |
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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>. | |
+ | <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 and , then , in terms of , is:
Solution
Using the change-of-base rule: and .
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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