Question

( 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} )

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

Solution