2003 AMC 10B Problems/Problem 9

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.

$\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}$ $25^{-2} = (5^2)^{-2} = 5^{-4}$ $5^{-4} = 5^{-\frac{12}{x}}$

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.

$\begin{align*}-4&=\frac{-12}{x}\\ -4x&=-12\\ x&=\boxed{\textbf{(B) \ } 3}\end{align*}$

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2003 AMC 10B Problems/Problem 9

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