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EXERCISE 6.1 · Q80

Q.Solve for xx: log⁡2x+log⁡4x+log⁡16x=214\log_2 x+\log_4 x+\log_{16}x=\dfrac{21}{4}.

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log⁡2x+log⁡4x+log⁡16x=214\log_2x+\log_4x+\log_{16}x=\dfrac{21}{4}.

Convert to base 22: log⁡4x=log⁡2xlog⁡24=log⁡2x2\log_4x=\dfrac{\log_2x}{\log_24}=\dfrac{\log_2x}{2}, and log⁡16x=log⁡2xlog⁡216=log⁡2x4\log_{16}x=\dfrac{\log_2x}{\log_216}=\dfrac{\log_2x}{4}. …

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