1961 AHSME Problems/Problem 9

Problem 9

Let $r$ be the result of doubling both the base and exponent of $a^b$ , and $b$ does not equal to $0$ . If $r$ equals the product of $a^b$ by $x^b$ , then $x$ equals:

$\textbf{(A)}\ a \qquad \textbf{(B)}\ 2a \qquad \textbf{(C)}\ 4a \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$

Solution

From the problem, $r = (2a)^{2b}$ , so $(2a)^{2b} = a^b \cdot x^b$ $(4a^2)^b = (ax)^b$ $4a^2 = ax$ $x = 4a$ Thus, the answer is $\boxed{\textbf{(C)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1961 AHSME Problems/Problem 9

Problem 9

Solution

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