# 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.

$$(\sqrt{2})^x = (\sqrt{3})^x$$

• Apply the $\log$ function you find $x=0$.
– user63181
Mar 20 '14 at 20:36
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• Try dividing by $(\sqrt{3})^x$... Mar 20 '14 at 20:38
• Don't forget that $\sqrt{(2)} = 2^{1/2}$.
– R R
Mar 20 '14 at 22:26

take log both sides of your equation to get $x$ log$\sqrt 2$=$x$ log$\sqrt 3$ and hence $x$(log $\sqrt 2$-log$\sqrt 3$)=$0$ $\Longrightarrow$ $x=0$.
Maybe the quickest way is to divide both sides by $\sqrt2^x$. Then you get $1=\sqrt{3/2}^x$, and you know that for a number $a>1$, the only power $a^x$ that’s equal to $1$ is for $x=0$.
Indeed the procedure that will generally allow you to solve equation of that sort explicitly uses logarithms, as it has been said. But that's a simple case: for two numbers $0\neq a\neq b\neq0$ and $a^x=b^x$, $x=1$ is ruled out. And because of that, so is every other number except $0$, the only number $x$ for which $\forall a\ \forall b, a^x=b^x$. So that's the solution. Just to make the point equations are not just puzzles to solve in according to mindless rules, but have a meaning, and thus various ways, or at least interpretations, on how to solve them.