1985 AHSME Problems/Problem 15

Problem

If $a$ and $b$ are positive numbers such that $a^b = b^a$ and $b = 9a$ , then the value of $a$ is

$\mathrm{(A)\ } 9 \qquad \mathrm{(B) \ }\frac{1}{9} \qquad \mathrm{(C) \ } \sqrt[9]{9} \qquad \mathrm{(D) \ } \sqrt[3]{9} \qquad \mathrm{(E) \ }\sqrt[4]{3}$

Solution

Substituting $b = 9a$ into $a^b = b^a$ gives $\begin{align*}a^{9a} = \left(9a\right)^a &\iff \left(a^9\right)^a = \left(9a\right)^a \qquad \text{(using the identity } \left(x^y\right)^z = x^{yz}\text{ for } x > 0\text{)} \\ &\iff a^9 = 9a \qquad \text{(taking the } a\text{th root of both sides, as } a > 0\text{)} \\ &\iff a^8 = 9 \qquad \text{(as } a \neq 0\text{)} \\ &\iff a^4 = 3 \qquad \text{(taking square roots and noting that } a^4 \geq 0\text{}) \\ &\iff a = \boxed{\text{(E)} \ \sqrt[4]{3}} \qquad \text{(again as } a > 0\text{)}.\end{align*}$

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1985 AHSME Problems/Problem 15

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