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Question Number 172077 by Mikenice last updated on 23/Jun/22
solve log_4 (x+12).logx^2 =1
solvelog4(x+12).logx2=1
Commented by mr W last updated on 24/Jun/22
log (x+12)log x^2 =log 4=2 log 2 −12<x<0: 2 log (x+12)log (−x)=2 log 2 log (x+12)log (−x)=log 2 log (−10+12)log (10)=log 2 ✓ log (−2+12)log (2)=log 2 ✓ x=−10, −2 are roots. x>0: log (x+12)log (x)=log 2 we can only approximate. x≈1.8358
log(x+12)logx2=log4=2log2−12<x<0:2log(x+12)log(−x)=2log2log(x+12)log(−x)=log2log(−10+12)log(10)=log2✓log(−2+12)log(2)=log2✓x=−10,−2areroots.x>0:log(x+12)log(x)=log2wecanonlyapproximate.x≈1.8358

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