Question

Solve the following inequalities.
$(1/2{)}^{lo{g}_2(x^2-2x-3)}>1.$

Answer 1

$(1/2{)}^{{\log }_2(x^2-2x-3)}$>1

For $\log$ to be defined-

$x^2-2x-3>0$

$⟹(x-3)(x+1)>0$

$⟹x<-1$ or $x>3$

Now from inequality-

$\frac{1}{2^{{\log }_2(x^2-2x-3)}}>1$

$⟹\frac{1}{x^2-2x-3}>1$

$⟹x^2-2x-3<1$

$⟹x^2-2x-4<0$

The solutions of equation $x^2-2x-4=0$ are $x=\frac{2\pm \sqrt{4+16}}{2}$

Hence, $1-\sqrt{5}<x<1+\sqrt{5}$

Hence final solution is-

$x\in (1-\sqrt{5},-1)\cup (3,1+\sqrt{5})$