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Formatted:

$$3^{\log_2 x} = x^{\log_2 3}$$

I know that $\log_2 x = \frac{\log_3 x}{\log_3 2}$ and that $3^{\log_3 x} = x$; but how would the $\frac{1}{\log_3 2}$ part of the exponent fit in?

What are the properties involved?

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    $\begingroup$ You may take the $log$ base $2$ on both sides. By one of the most well known log properties, you get on both sides of the equal sign a product of.....two equal terms $\endgroup$ – imranfat May 22 '18 at 20:45
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Continuing where you left off just for the sake of it (taking $\log_2$ of both sides is a lot simpler), we currently have:

$$x^{1/\log_3(2)} = x^{\log_2(3)}$$

We know that:

$${1 \over \log_3(2)} = {\log_3(3) \over \log_3(2)} = \log_2(3)$$

Which gives the desired exponent.

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It is not necessary to change base. You just take the $\log_2$ of both sides and apply the properties of logarithms. You get $$\log_2 3\cdot \log_2 x = \log_2 3\cdot \log_2 x,$$ which is true for any $x > 0$.

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\begin{eqnarray*} 3^{\log_2(x)}=2^{\log_2(3^{\log_2(x)})}= 2^{\log_2(3) \log_2(x)}=2^{\log_2(x^{\log_2(3)})}=x^{\log_2(3)}. \end{eqnarray*}

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  • $\begingroup$ There should be no exponent in exponents. $\endgroup$ – Bernard May 22 '18 at 20:55
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original equation: $$3^{\log_2 x} = x^{\log_2 3}$$ take $\log_2$ of both sides: $$\log_2 (3^{\log_2 x}) = \log_2 (x^{\log_2 3})$$ use exponent rules: $$\log_2 x \log_2 3 = \log_2 3 \log_2 x$$ we know that $\log_2 3 = \log_2 3$, and that $\log_2 x = \log_2 x$; we also know that $xy = xy$; thus we know that the above is true.

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