Difference between revisions of "1982 AHSME Problems/Problem 13"
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+ | == Problem 13 == | ||
+ | |||
+ | If <math>a>1, b>1</math>, and <math>p=\frac{\log_b(\log_ba)}{\log_ba}</math>, then <math>a^p</math> equals | ||
+ | |||
+ | <math>\textbf {(A)}\ 1 \qquad | ||
+ | \textbf {(B)}\ b \qquad | ||
+ | \textbf {(C)}\ \log_ab \qquad | ||
+ | \textbf {(D)}\ \log_ba \qquad | ||
+ | \textbf {(E)}\ a^{\log_ba} </math> | ||
+ | |||
+ | ==Solution 1== | ||
p (log <sub> b</sub> a) = log <sub>b </sub> (log <sub> b</sub> a) | p (log <sub> b</sub> a) = log <sub>b </sub> (log <sub> b</sub> a) | ||
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a<sup>p</sup> = log <sub> b</sub> a | a<sup>p</sup> = log <sub> b</sub> a | ||
+ | |||
+ | ==Solution 2== | ||
+ | Notice that <math>\frac{\log_b(\log_ba)}{\log_ba}</math> strongly resembles the chain rule. Recall that <math>\log_ba=\frac{\log_ca}{\log_cb}</math>. Taking the base on the RHS to be <math>b</math>, we get that <math>p = \log_a(\log_ba)</math>. Raising <math>a</math> to both sides, we get that <cmath>a^p = a^{\log_a(\log_ba)}</cmath> <cmath>= \boxed{\textbf{(D)} ~ \log_ba}</cmath> | ||
+ | |||
+ | ~ cxsmi |
Revision as of 00:13, 13 May 2024
Problem 13
If , and , then equals
Solution 1
p (log b a) = log b (log b a)
log b (a p) =log b (logb a)
ap = log b a
Solution 2
Notice that strongly resembles the chain rule. Recall that . Taking the base on the RHS to be , we get that . Raising to both sides, we get that
~ cxsmi