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How can I solve the following equation:

$$2^{log_3 x}+x^{log_3 2}=4$$

I don't want the final answer, I want to know how I can solve these kind of equations.

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At least the notation $log(x,3)$ is not standard in maths. Do you mean $\log_x 3$ or $\log_3 x$? –  Jaakko Seppälä Sep 9 '12 at 10:35
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You should probably write what you mean by $\log (x,3)$. Or better yet, use standard convention by writing either $\log_x(3)$ or $\log_3(x)$. –  tomasz Sep 9 '12 at 10:35
    
There isn't a way to solve all equations, but often you should start by writing everything you can as an exponent of the same base. In this case, notice what happens if you changed $2$ to $3^{\log_3 2}$ and $x$ to $3^{\log_3 x}$. –  Karolis Juodelė Sep 9 '12 at 10:57
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1 Answer

up vote 4 down vote accepted

Put $x=3^y$

Simplify $2^{log_3 x}$ and $x^{log_3 2}$ as follows:

$log_3 x=log_3 3^y=ylog_3 3=y$

$x^{log_3 2}=(3^y)^{log_3 2}=(3^{log_32})^y=2^y$

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