jmc

We can start by cross-multiplying: $$\begin{array}{rl}3\sqrt{x-1}& =2\sqrt{x}\\ \Rightarrow (3\sqrt{x-1}{)}^2& =(2\sqrt{x}{)}^2\\ \Rightarrow 9(x-1)& =4(x)\\ \Rightarrow 9x-9& =4x\\ \Rightarrow 5x& =9\\ \Rightarrow x& =\boxed{\frac{9}{5}}.\end{array}$$Checking, we see that this value of $x$ does indeed work, so it is not an extraneous solution.