What are the steps in showing $3^{ \log_2 x} = x^{\log_2 3}$

Question

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?

Answer 1

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.

Answer 2

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 $x > 0$.

Answer 3

\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*}

Answer 4

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.