Explain the logic in this math equation

Question

$$ n = 4^x \\ n = 2^x \times 2^x \\ \sqrt n = 2^x \\ \log_2\sqrt n = x $$

I do not understand how $$ n = 2^x \times 2^x $$ is transformed into $$ \sqrt n = 2^x $$

Square root is undone by raising something to the 2nd power. But they somehow managed to undo it.

Please ELI5.

Thank you.

Answer 1

The trick is behind transforming the number $4$.

Basically, the original equation can be rewritten as follows:

$$ n = 4^x \\ n = (2^2)^x $$

Using the properties of exponents, we can rewrite $(2^2)^x$ as $(2^x)^2$. Then, taking the square root of both sides, we get:

$$ n = (2^x)^2 \\ \sqrt{n} = 2^x $$

Another way to think about it is: that in the second step of your original process, $n$ is equal to $2^x \cdot 2^x$. Anything multiplied by itself is that thing squared, so $2^x \cdot 2^x = (2^x)^2$. Now, we can take the square root again:

$$ n = (2^x)^2 \\ \sqrt{n} = 2^x $$

Answer 2

Let $u=2^x$. The step you are confused about becomes $n=u*u=u^2 \implies \sqrt{n}=u$ For positive $u$, this is true. Like you said, the square root is the inverse function to squaring.