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

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==Solution 2==
 
==Solution 2==
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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>
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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>
  
 
~ cxsmi
 
~ cxsmi
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==See Also==
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{{AHSME box|year=1982|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 23:11, 29 June 2025

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

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
All AHSME Problems and Solutions


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