Question

(log x) 08 ** r) log²x+ log x2-2 = log Vx. ( 2

JEE/Engineering Exams

Maths

Solution

149

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( begin{aligned}left(frac{log x}{2}right)^{log _{. x}+log x^{2}-2} &=log sqrt{x} &=frac{1 log }{2} &=left(frac{log x}{2}right)^{frac{1}{2}} end{aligned} ) so exponents thould be equal ( log _{x}^{2}+log x^{2}-2=1 ) ( log ^{2} x+2 log x-3 leq 0 ) ( log _{x} y= ) ( y^{2}+2 x-3=0 ) ( y^{2}+3 y-y-3=0 ) ( y(y+3)-(y+3)=0 ) ( (y-1)(y+3)=0 ) ( y=1,-8 ) ( x=10, frac{1}{1000} )