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Difference between revisions of "1982 AHSME Problems/Problem 13"

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== Problem ==
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If <math>a>1, b>1</math>, and <math>p=\frac{\log_b(\log_ba)}{\log_ba}</math>, then <math>a^p</math> equals
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<math>\textbf {(A)}\ 1 \qquad
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\textbf {(B)}\ b \qquad
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\textbf {(C)}\ \log_ab \qquad
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\textbf {(D)}\ \log_ba \qquad
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\textbf {(E)}\ a^{\log_ba} </math>
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==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
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==Solution 2==
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Notice that <math>\frac{\log_b(\log_ba)}{\log_ba}</math> strongly resembles the change of base 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>
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~ cxsmi

Latest revision as of 10:07, 13 May 2024

Problem

If $a>1, b>1$, and $p=\frac{\log_b(\log_ba)}{\log_ba}$, then $a^p$ equals

$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ b \qquad \textbf {(C)}\ \log_ab \qquad \textbf {(D)}\ \log_ba \qquad \textbf {(E)}\ a^{\log_ba}$

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 $\frac{\log_b(\log_ba)}{\log_ba}$ strongly resembles the change of base rule. Recall that $\log_ba=\frac{\log_ca}{\log_cb}$. Taking the base on the RHS to be $b$, we get that $p = \log_a(\log_ba)$. Raising $a$ to both sides, we get that \[a^p = a^{\log_a(\log_ba)}\] \[= \boxed{\textbf{(D)} ~ \log_ba}\]

~ cxsmi

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