#### Answer

The solution set is $\left\{2\right\}$.

#### Work Step by Step

Square both sides to obtain:
$x^2=(2\sqrt{x-1})^2$
Use the rule $(ab)^m = a^mb^m$ to obtain:
$x^2=2^2(\sqrt{x-1})^2
\\x^2=4(x-1)
\\x^2=4x-4$
Subtract $4x$ and add $4$ on both sides of the equation to obtain:
$x^2-4x+4 = 4x-4-4x+4
\\x^2-4x+4=0$
Factor the trinomial to obtain:
$(x-2)(x-2) = 0$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&x-2=0 &\text{ or } &x-2=0
\\&x=2 &\text{ or } &x=2
\end{array}$
Thus, the solution set is $\left\{2\right\}$.