Solving ${\sqrt2}^{\,x} = {\sqrt3}^{\,x}$ I am studying logarithms and exponents.  I am not sure how to go about solving this problem.  I seem too keep going in circles using the rules of log and exp.
$$(\sqrt{2})^x = (\sqrt{3})^x$$

take log both sides of your equation to get $x$ log$\sqrt 2$=$x$ log$\sqrt 3$ and hence $x$(log $\sqrt 2$-log$\sqrt 3$)=$0$ $\Longrightarrow$ $x=0$.
