How does $2^{(\log_4{x})}$ become $\sqrt[2]{x}$?

I'm having trouble of figuring out the steps to simplify the expression below, I have the answer but no clue how to reach this.

$$2^{(\log_4{x})} = \ \cdots \ = \sqrt[2]{x}$$

Update, thanks to the first comment I found the solution:

$2^{\log_{4}{x}} = \big( 4^{\log_{4}{x}} \big )^{1/2} = x^{1/2} = \sqrt{x}$

Thanks a lot! I wouldn't have figured it out otherwise.

Solution 1:

You should begin by making a logarithm change of base. Suppose you have base "a" and want to change to base "c", overall the process is like this:

$$\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$$

In this case you have $\log_4 (x)$ and want to write it as $\log_2(x)$ so $$\log_4(x) = \frac{\log_2(x)}{\log_2(4)}$$ and $\log_2(4) = 2$ so $$\log_4(x) = \frac{1}{2}\log_2(x)$$

Now you have

$$2^{\log_4(x)} = 2^{\frac{1}{2}\log_2(x)} = (2^{\log_2(x)})^{\frac{1}{2}} = x^{\frac{1}{2}} = \sqrt x$$

Solution 2:

In general $\Large N = B^{\log_B (N)}$ and $\Large (B^n)^m = B^{n\times m}$

$\Large \sqrt x = x^{1/2} = (4^{\log_4 (x)})^{1/2}= ((2^2)^{\log_4(x)})^{1/2}=(2^{2\times\log_4(x)})^{1/2}=2^{2\times 1/2\times \log_4(x)}= 2^{\log_4(x)}$.

The left to right direction explains how $\sqrt x$ becomes $2^{\log_4(x)}$, the right to left direction, which is more difficult to find, explains the reverse process.