Question

Solve the following equations:
$\sqrt{2^x}+\frac{1}{\sqrt{2^x}}=2$.

Answer 1

$\sqrt{2^x}+\frac{1}{\sqrt{2^x}}=2$

Put $\sqrt{2^x}=t$

$t+\frac{1}{t}=2\frac{t^2+1}{t}=2t^2+1=2t{t}^2-2t+1=0{(t-1)}^2=0t=1\sqrt{2^x}=t{2}^x=t^2{2}^x=1^2=2^0\Rightarrow x=0$