2003 AMC 10B Problems/Problem 9

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Problem

Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$

$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 9$

Solution

Manipulate the powers of $5$ in order to get a clean expression.

$25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}$

$(5^2)^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot (5^2)^{17/x}}$

$5^{-4} = \frac{5^{48/x}}{5^{26/x} \cdot 5^{34/x}}$

$5^{-4} = 5^{48/x-26/x-34/x}$

$5^{-4} = 5^{(48-26-34)/x}$

$5^{-4} = 5^{-12/x}$

If two numbers are equal, and their bases are equal, then their exponents are equal as well. Set the two exponents equal to each other.

$-4=\frac{-12}{x}$

$-4x=-12$

$\boxed{x=3 \text{ (B)}}$

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 10 Problems and Solutions