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 $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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