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.

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

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    $\begingroup$ Apply the $\log $ function you find $x=0$. $\endgroup$
    – user63181
    Mar 20 '14 at 20:36
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    $\begingroup$ Questions regarding homework assignments are more than welcome, provided that they: Briefly explain the problem you are trying to solve—do not post your entire assignment verbatim. Explain what you tried and where you're stuck (showing your work is a good idea). Don't ask for complete solutions to the problem—we're not here to do your homework for you. $\endgroup$ Mar 20 '14 at 20:37
  • $\begingroup$ Try dividing by $(\sqrt{3})^x$... $\endgroup$
    – geodude
    Mar 20 '14 at 20:38
  • $\begingroup$ Don't forget that $\sqrt{(2)} = 2^{1/2}$. $\endgroup$
    – 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$\sqrt 3$)=$0$ $\Longrightarrow$ $x=0$.

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    $\begingroup$ This answer helped me understand the material more. The whole subject was confusing me a little, but now is more clear. $\endgroup$ Mar 20 '14 at 20:51
  • $\begingroup$ good to know that $\endgroup$
    – wanderer
    Mar 20 '14 at 20:52

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.


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