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

m
Line 1: Line 1:
 +
== 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)
  
Line 4: Line 15:
  
 
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 $a>1, b>1$, 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