Question

Solve: ${\left(\frac{1}{2}\right)}^{\log x^2}+2>{3.2}^{-\log (-x)}$

Answer 1

$(\frac{1}{2}{)}^{\log x^2}+2>{3.2}^{-\log (-x)}$

$\Rightarrow 2^{-\log x^2}+2>{3.2}^{-\log (-x)}$

This can be wtitten as

$\Rightarrow 2^{-\log (-x{)}^2}+2>{3.2}^{-\log (-x)}$

Take $2^{-\log (-x)}=t$

$\Rightarrow t^2+2>3t$

$\Rightarrow (t-1)(t-2)>0$

$\Rightarrow t<1;t>2$

$\Rightarrow 2^{-\log (-x)}<1$ ; $\Rightarrow 2^{-\log (-x)}>2$

$\Rightarrow -\log (-x)<2^0$ ; $\Rightarrow 2^{-\log (-x)}>2^1$

$\Rightarrow -\log (-x)<0$ ; $\Rightarrow -\log (-x)>1$

$\Rightarrow \log (-x)>0$ ; $\Rightarrow \log (-x)<-1$

$\Rightarrow -x>{10}^0$ ; $\Rightarrow -x<{10}^{-1}$

$\Rightarrow -x>1$ ; $\Rightarrow -x<0.1$

$\Rightarrow x<-1$ ; $\Rightarrow x>-0.1$

Also for the logarithm to exist we must have $-x>0\Rightarrow x<0$

Therefore $x$ lies in the interval $(-\infty ,-1)\cup (-0.1,0)$