Question

Solve for x: ${\log }^2(4-x)+\log (4-x).\log (x+\frac{1}{2})-2{\log }^2(x+\frac{1}{2})=0$

Answer 1

$Wecansaythat$

$\log {}^2(4-x)+\log (4-x)=0and\log (x+\frac{1}{2})-2\log {}^2(x+\frac{1}{2})=0$

$\log (4-x)=0$

$4-x=1$

$x=3$

$and\log (4-x)=-1$

$4-x=e^{-1}$

$x=4-e^{-1}$

$Now,\log (x+\frac{1}{2})-2\log {}^2(x+\frac{1}{2})=0$

$4-x=e^{-1}$

$x=4-e^{-1}$

$Now,\log (x+\frac{1}{2})-2\log {}^2(x+\frac{1}{2})=0$

$So,\log (x+\frac{1}{2})=0$

$x+\frac{1}{2}=1$

$x=\frac{1}{2}$

$or$

$1-2\log (x+\frac{1}{2})=0$

$\log (x+\frac{1}{2})=\frac{1}{2}$

$x+\frac{1}{2}=e^{\frac{1}{2}}$

$x=\sqrt{e}-\frac{1}{2}$

$thus,x=\frac{1}{2},4-\frac{1}{e},\sqrt{e}-\frac{1}{2}$