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

Difference between revisions of "1982 AHSME Problems/Problem 13"
m (→‎Solution 2)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
−
== Problem 13 ==
+
== 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  
Line 7: Line 7:
 
\textbf {(C)}\ \log_ab \qquad  
 
\textbf {(C)}\ \log_ab \qquad  
 
\textbf {(D)}\ \log_ba \qquad  
 
\textbf {(D)}\ \log_ba \qquad  
−
\textbf {(E)}\ a^{\log_ba} </math>  
+
\textbf {(E)}\ a^{\log_ba} </math>
  
 
==Solution 1==
 
==Solution 1==
Line 17: Line 17:
  
 
==Solution 2==
 
==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>
+
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_13&oldid=219284"