1982 AHSME Problems/Problem 13

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 (log_b a)

a^p = 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

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1982 AHSME Problems/Problem 13

Problem

Solution 1

Solution 2