Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.9 AMC 10/12 Find the answer China

If real number $x$ satisfies $\log_2 x = \log_4(2x) + \log_8(4x)$ , then the value of $x$ is ______.

Your answer

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

Solution

By the given condition, we have
$\log_2 x = \log_4 2 + \log_4 x + \log_8 4 + \log_8 x = \frac{1}{2} + \frac{1}{2} \log_2 x + \frac{2}{3} + \frac{1}{3} \log_2 x,$
and its solution is $\log_2 x = 7$ . Therefore, $x = 128$ .

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