Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.1 AIME Find the answer United States

Problem:
How many real numbers $x$ are solutions to the following equation?
$2003^{x} + 2004^{x} = 2005^{x}$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Rewrite the equation as $(2003 / 2005)^{x} + (2004 / 2005)^{x} = 1$ . The left side is strictly decreasing in $x$ , so there cannot be more than one solution. On the other hand, the left side equals $2 > 1$ when $x = 0$ and goes to $0$ when $x$ is very large, so it must equal $1$ somewhere in between. Therefore there is one solution.

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