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Difference between revisions of "1982 AHSME Problems/Problem 13"
m (→‎Problem 13)
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== Problem 13 ==
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== Problem ==
  
 
If <math>a>1, b>1</math>, and <math>p=\frac{\log_b(\log_ba)}{\log_ba}</math>, then <math>a^p</math> equals  
 
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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\textbf {(C)}\ \log_ab \qquad  
 
\textbf {(C)}\ \log_ab \qquad  
 
\textbf {(D)}\ \log_ba \qquad  
 
\textbf {(D)}\ \log_ba \qquad  
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\textbf {(E)}\ a^{\log_ba} </math>  
+
\textbf {(E)}\ a^{\log_ba} </math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 11:07, 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 chain 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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