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

Revision as of 00:13, 13 May 2024

Problem 13

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 chain 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

@@ Line 1: / Line 1: @@
+== Problem 13 ==
+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)
@@ Line 4: / Line 15: @@
 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 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>
+~ cxsmi

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

Revision as of 00:13, 13 May 2024

Problem 13

Solution 1

Solution 2