<s>[INST] Solve the inequality $(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$ 
Solve the inequality $$(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$$

My work so far:
1) $a>0$
2) Let $f(x)=(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)$. 
$f(x)=0$ if and only if $x^2-2\sqrt{a}\cdot x+1=0$ or $2^x+\lg a=0$
$$x^2-2\sqrt{a}\cdot x+1=0$$
$$x_{1,2}=\sqrt a \pm \sqrt{a-1}$$
and $$2^x+\lg a=0$$
$$x=\log_2\lg\frac1a$$
 [/INST] $1)$ If $0<a<1$ then we have $x^2-2\sqrt{a}\cdot x+1>0$ for all real number $x$. 
Then we must have 
$$2^x+\log a<0\to 2^x<\log a^{-1}\to x<\log_2(\log a^{-1})$$
$2)$ If $a>1$ then we have $2^x+\log a>0$ for all real number $x$.
Then we must have 
$$x^2-2\sqrt{a}\cdot x+1<0\to \sqrt{a}-\sqrt{a-1}<x<\sqrt{a}+\sqrt{a-1}$$
$3)$ If $a=1$ then
$$(x^2-2x+1)(2)=2(x-1)^2\ge0$$
and then we have no solution.
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