Solve: 2 log2(log2x) + log1/2(log2sq.root2x)=1

Dear Student,
Please find below the solution to the asked query:

2log2log2x+log12log222x=12log2log2x+log2-1log222x=1Using identities.logamn=n×logamloganm=1nlogam, we get,log2log2x2-log2log222x=1Using identitylogamn=logam-logan, we getlog2log2x2log22x=1log2x2log22x=21log2x2=2log222xlog2x2=2log222xlog2x2=2log28xlog2x2=2log2x+log28log2x2=2log2x12+log2232log2x2=212log2x+32log22We know that logaa=1log2x2=log2x+3log2x2-log2x-3=0which is quadratic in log2xlog2x=--1±-12-4×1×-32log2x=1±132log2x=1+132 or log2x=1-132x=21+132 or x=21-132

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