AMC 10 Step by Step

2003 AMC 10B Problem 9

P1-10~2 minPrint

Find the value of xx that satisfies the equation 25−2=548/x526/x⋅2517/x.25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.

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Key insight

Write every number as a power of 5. Then the exponents must match: -4 = -12/x, so x = 3.

Solution

1. Everything here is built from 55, since 25=5225 = 5^2. So rewrite every piece as a power of 55.
2. Left side: 25−2=(52)−2=5−425^{-2} = (5^2)^{-2} = 5^{-4}.
3. The 2525 on the bottom right: 2517/x=(52)17/x=534/x25^{17/x} = (5^2)^{17/x} = 5^{34/x}.
4. Now the right side is
548/x526/x⋅534/x.\frac{5^{48/x}}{5^{26/x} \cdot 5^{34/x}}.
Multiplying powers of 55 adds the exponents, and dividing subtracts them. So the exponent is 48x−26x−34x=48−26−34x=−12x\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/x5^{-4} = 5^{-12/x}. Two powers of 55 are equal only when their exponents are equal:
−4=−12x,sox=3.-4 = -\frac{12}{x}, \quad\text{so}\quad x = 3.

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

Why this works

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 2525 is 525^2, turn everything into base 55, and never try to compute the huge numbers themselves.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Convert every factor to base 55. Since 25=5225 = 5^2,
25−2=5−4,2517/x=534/x.25^{-2} = 5^{-4}, \qquad 25^{17/x} = 5^{34/x}.

The right side is then a single power of 55 whose exponent is the top exponent minus the two bottom exponents:
548/x526/x⋅534/x=5(48−26−34)/x=5−12/x.\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=−12x⟹x=3.-4 = -\frac{12}{x} \quad\Longrightarrow\quad x = 3.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.