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

algebra-precalculuslogarithms

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.

Best Answer

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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