Question

Solve the following equation for x
$x^{\left[\frac{3}{4}\right({\log }_{2}x{)}^2+{\log }_{2}x-\frac{5}{4}]}=\sqrt{2}$

Answer 1

Given:

$x^{\left[\frac{3}{4}\right({\log }_{2}x{)}^2+{\log }_{2}x-\frac{5}{4}]}=\sqrt{2}$

taking ${\log }_{2}x$ on both sides,

$\left[\frac{3}{4}\right({\log }_{2}x{)}^2+{\log }_{2}x-\frac{5}{4}]{\log }_{2}x={\log }_2\sqrt{2}$

Put ${\log }_{2}x=t$

$\Rightarrow [\frac{3}{4}{t}^2+t-\frac{5}{4}]t=\frac{1}{2}{\log }_{2}2$

$3t^3+4t^2-5t=2$

$3t^3+4t^2-5t-2=0$

$\Rightarrow 3t^3-3t^2+7t^2-7t+2t-2=0$

$\Rightarrow 3t(t-1)++7t(t-1)+2(t-1)=0$
$\Rightarrow (t-1)(3t^3+7t+2)=0$
$\Rightarrow (t-1)(t+2)(3t+1)=0$
$t={\log }_{2}x=1,-2,-\frac{1}{3}$

${\log }_{2}x=1$

$\Rightarrow x=2$

${\log }_{2}x=-2$

$\Rightarrow x=-\frac{1}{4}$

${\log }_{2}x=-\frac{1}{3}$

$\Rightarrow x=\frac{1}{2^{\frac{1}{3}}}$

$\therefore ,x=2,-\frac{1}{3},\frac{1}{2^{\frac{1}{3}}}$