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Exercise 2.12 · Q11

Q.Solve log⁡2x−3log⁡1/2x=6\log_2x-3\log_{1/2}x=6.

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Step 1. log⁡1/2x=log⁡xlog⁡(1/2)=log⁡x−log⁡2=−log⁡2x\log_{1/2}x=\dfrac{\log x}{\log(1/2)}=\dfrac{\log x}{-\log2}=-\log_2x.

Step 2. So −3log⁡1/2x=−3(−log⁡2x)=3log⁡2x-3\log_{1/2}x=-3(-\log_2x)=3\log_2x. …

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