<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] you must do case work for $a$: 
1) $$a>1$$ then we have $$2^x+\lg(a)>0$$ and we have to solve $$x^2-2\sqrt{a}x+1<0$$  this is equivalent to
$$(x-\sqrt{a})^2<a-1$$ and further
$$|x-\sqrt{a}|<\sqrt{a-1}$$
can you proceed?
2)$$a=1$$ then we have $$(x^2-2x+1)<0$$ which is impossible.
3) the case $$0<a<1$$ is for you!
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