Question

Solve the following inequality:
${\log }_2(x+14)+2{\log }_4(x+2)<2{\log }_{1/2}\left(\frac{1}{8}\right)$

Answer 1

${\log }_2(x+14)+2{\log }_4(x+2)<2{\log }_{1/2}\left(\frac{1}{8}\right)$

${\log }_2(x+14)+{\log }_2(x+2)<2{\log }_2\left(8\right)$

${\log }_2(x+14)(x+2)<{\log }_2\left(64\right)$

$(x+14)(x+2)<64$

$x^2+16x+28-64<0$

$x^2+16x-36<0$

$(x+18)(x-2)<0$

$xϵ(-18,2)$

Also for

$\log (x+14)\log (x+2)$

$(x+14)>0x+2>0$

$x>-14x>-2$

$xϵ(-14,\infty )xϵ(-2,\infty )$

Taking intersection we get $xϵ(-2,2)$