Question

Solve the following inequality:
${\left(\frac{1}{3}\right)}^{{\log }_{1/3}(3.(1/2{)}^x+5)}<2^{1+x}$

Answer 1

$(\frac{1}{3}{)}^{{log}_{\frac{1}{3}}(3.{\left(\frac{1}{2}\right)}^x+5)}<2^{1+x}\Rightarrow 3.{\left(\frac{1}{2}\right)}^x+5<2^{1+x}$

Let $2^x=a\Rightarrow a>0$

Therefore, $\frac{3}{a}+5<2a3+5a<{2a}^2{2a}^2-5a-3>0(2a+1)(a-3)>0a\in (3,\infty )(\because a>0,2a+1>0(always\left)\right)x\in ({log}_{2}3,\infty ).$