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Algebra Difficulty 4.4 AIME Prove it United States

Problem:
A real number $x$ satisfies $9^{x} + 3^{x} = 6$ . Compute the value of $16^{1 / x} + 4^{1 / x}$ .

Solution

Solution:
Setting $y = 3^{x}$ in the given equation yields

$y^{2} + y = 6 \Longrightarrow y^{2} + y - 6 = 0 \Longrightarrow y = -3, 2$

Since $y > 0$ we must have

$3^{x} = 2 \Longrightarrow x = \log_{3}(2) \Longrightarrow 1 / x = \log_{2}(3)$

This means that

$16^{1 / x} + 4^{1 / x} = \left(2^{1 / x}\right)^{4} + \left(2^{1 / x}\right)^{2} = 3^{4} + 3^{2} = 90.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.