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Algebra Difficulty 5.3 AIME, harder Prove it Slovenia

Find all real numbers xx that solve the equation
log2(10x)+log4(100x)+log8(1000x)2log64x=9. \log_2(10x) + \log_4(100x) + \log_8(1000x) - 2\log_{64} x = 9.
Write the result as a reduced fraction.

Solution

Rewrite the equation as
9=log2(10x)+log4(100x)+log8(1000x)2log64x=log2(10x)+log(100x)log4+log(1000x)log82logxlog64=log2(10x)+log(100x)2log2+log(1000x)3log22logx6log2=log2(10x)+12log2(100x)+13log2(1000x)13log2x=log2(10x)+log2(100x)+log2(1000x3)log2x3=log2(10x100x1000x3x3)=log2(1000x33). \begin{align*} 9 &= \log_2(10x) + \log_4(100x) + \log_8(1000x) - 2\log_{64} x \\ &= \log_2(10x) + \frac{\log(100x)}{\log 4} + \frac{\log(1000x)}{\log 8} - \frac{2\log x}{\log 64} \\ &= \log_2(10x) + \frac{\log(100x)}{2\log 2} + \frac{\log(1000x)}{3\log 2} - \frac{2\log x}{6\log 2} \\ &= \log_2(10x) + \frac{1}{2}\log_2(100x) + \frac{1}{3}\log_2(1000x) - \frac{1}{3}\log_2 x \\ &= \log_2(10x) + \log_2(\sqrt{100x}) + \log_2(\sqrt[3]{1000x}) - \log_2\sqrt[3]{x} \\ &= \log_2\left(\frac{10x \cdot \sqrt{100x} \cdot \sqrt[3]{1000x}}{\sqrt[3]{x}}\right) \\ &= \log_2(1000\sqrt[3]{x^3}). \end{align*}
It follows that 1000x33=291000\sqrt[3]{x^3} = 2^9 or x=(29103)23=1625x = \left(\frac{2^9}{10^3}\right)^{\frac{2}{3}} = \frac{16}{25}.

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