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Question Number 172077 by Mikenice last updated on 23/Jun/22

$$solve$$$${log}_4(x+12).{logx}^2=1$$

Commented by mr W last updated on 24/Jun/22

$$\log (x+12)\log x^2=\log 4=2\log 2$$$$-12<x<0:$$$$2\log (x+12)\log (-x)=2\log 2$$$$\log (x+12)\log (-x)=\log 2$$$$\log (-10+12)\log \left(10\right)=\log 2✓$$$$\log (-2+12)\log \left(2\right)=\log 2✓$$$$x=-10,-2areroots.$$$$x>0:$$$$\log (x+12)\log \left(x\right)=\log 2$$$$wecanonlyapproximate.x\approx 1.8358$$

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