2003 AMC 10B Problem 9

1. Everything here is built from $5$, since $25 = 5^2$. So rewrite every piece as a power of $5$.
2. Left side: $25^{-2} = (5^2)^{-2} = 5^{-4}$.
3. The $25$ on the bottom right: $25^{17/x} = (5^2)^{17/x} = 5^{34/x}$.
4. Now the right side is
$$\frac{5^{48/x}}{5^{26/x} \cdot 5^{34/x}}.$$
Multiplying powers of $5$ adds the exponents, and dividing subtracts them. So the exponent is $\frac{48}{x} - \frac{26}{x} - \frac{34}{x} = \frac{48 - 26 - 34}{x} = -\frac{12}{x}$.
5. The equation is now $5^{-4} = 5^{-12/x}$. Two powers of $5$ are equal only when their exponents are equal:
$$-4 = -\frac{12}{x}, \quad\text{so}\quad x = 3.$$

The answer is $\boxed{\textbf{(B)}\ 3}$.

When every number in an equation is a power of the same base, the equation is really about the exponents, and those follow simple adding and subtracting rules. Spot that $25$ is $5^2$, turn everything into base $5$, and never try to compute the huge numbers themselves.

Convert every factor to base $5$. Since $25 = 5^2$,
$$25^{-2} = 5^{-4}, \qquad 25^{17/x} = 5^{34/x}.$$

The right side is then a single power of $5$ whose exponent is the top exponent minus the two bottom exponents:
$$\frac{5^{48/x}}{5^{26/x} \cdot 5^{34/x}} = 5^{(48 - 26 - 34)/x} = 5^{-12/x}.$$

Equal powers of the same base have equal exponents, so
$$-4 = -\frac{12}{x} \quad\Longrightarrow\quad x = 3.$$

The answer is $\boxed{\textbf{(B)}\ 3}$.