Difference between revisions of "1982 AHSME Problems/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 | ||
| + | |||
| + | <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) | ||
| − | log <sub> b</sub> (a<sup> p</sup> =log <sub> b </sub> (log<sub>b</sub> a) | + | |
| + | log <sub> b</sub> (a<sup> p</sup>) =log <sub> b </sub> (log<sub>b</sub> a) | ||
| + | |||
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 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 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1982|num-b=12|num-a=14}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 23:11, 29 June 2025
Contents
Problem
If 
, and 
, then 
equals
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 
strongly resembles the change of base rule. Recall that 
. Taking the base on the RHS to be 
, we get that 
. Raising 
to both sides, we get that ![\[a^p = a^{\log_a(\log_ba)}\]](http://latex.artofproblemsolving.com/3/6/c/36c166c6678facc78a5e78037919032ce5eaa7c9.png)
![\[= \boxed{\textbf{(D)} ~ \log_ba}\]](http://latex.artofproblemsolving.com/d/9/0/d902082d18efea9acdcbaf6855b16eedbc4265be.png)
~ 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. 
