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

Substitue $b=9a$ into $a^b=b^a$ to get $a^{9a}=(9a)^a$. Since $x^{yz}=(x^y)^z$, we have $a^{9a}=(a^9)^a$, and $(a^9)^a=(9a)^a$. Taking the

$a\text{th}$ root of both sides gives $a^9=9a$. Dividing by $a$ yields $a^8=9\implies a=\sqrt[8]{9}=\sqrt[4]{3}, \boxed{\text{E}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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