AMC10 2003 10B, Problem 9

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

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&=\fbox{3}\end{align}

Answer: 3