Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

m (→‎Solution 2)
(→‎Solution 2)
 
Line 20: Line 20:
  
 
~ cxsmi
 
~ cxsmi
 +
 +
==See Also==
 +
{{AHSME box|year=1982|num-b=12|num-a=14}}
 +
 +
{{MAA Notice}}

Latest revision as of 22:11, 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


These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_13&oldid=252879"