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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.

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    $\begingroup$ Hint : $4^{\log_4(x)}=x$ since the functions $4^x$ and $\log_4(x)$ are inverse. $\endgroup$
    – Peter
    Nov 20, 2021 at 15:52
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    $\begingroup$ $2^{\log_4x} = 2^{\log_{2^2}x} = 2^{\frac{1}{2}\log_2x} = \sqrt{2^{\log_2x}} = \sqrt x$ $\endgroup$
    – VIVID
    Nov 20, 2021 at 15:59
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    $\begingroup$ I would start with $\log_4 x = \frac {\log_2 x}{\log_2 4} = \frac 12 \log_2 x$ $\endgroup$
    – user317176
    Nov 20, 2021 at 16:00

2 Answers 2

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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

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

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  • $\begingroup$ Didn't see you already add figured out the solution. Anyway I'Il keep the comment :) $\endgroup$
    – Dawn2011900
    Nov 20, 2021 at 16:18
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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.

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