Why the inverse function of $2^x $ is log and not root? I'm sure this is a very basic question, but while reading about functions and their inverse I'm having a hard time understanding why for $x^2$ or $x^3$ the inverse are $\sqrt x$ and $\sqrt[3]x$, but for $2^x$ is $\log_2 x$.
 A: Recall that by definition of inverse function, we must have that:
$$f\circ f^{-1}(x) = id(x) = x$$
Define
$f(x) = 2^x$ ,
$g(x) = \sqrt{x}$
Clearly: $f\circ g(x) = f(\sqrt{x}) = 2^{\sqrt{x}} \neq x$
Take, for example $x = 64$. Then we have $2^{\sqrt{64}} = 2^8 = 256 \neq 64$
A: There's two separate questions here:


*

*Why is $\log_2x$ the inverse of $2^x$?

*Why isn't $\sqrt{x}$ the inverse of $2^x$?
The answer to the first one is just the definition of logarithm: $\log_2x$ means "The number $2$ must be raised to to get $x$," so $2^{\log_2x}$ means "$2$ raised to (the number $2$ must be raised to to get $x$)," which is clearly $x$ itself. (Exercise: convince yourself that also $\log_2(2^x)=x$.) Basically, the logarithm is defined to be the inverse of $2^x$.
For an example of how to use this definition, let's say you were asked to calculate $\log_28$. You'd ask yourself, "What do I need to raise $2$ to to get $8$?," and quickly see that the answer is "$3$" - so $\log_28=3$. Now obviously understanding what $\log_2(\pi)$ is is going to be much more complicated, but the basic idea is the same.
OK,now what about $\sqrt{x}$? It's easy to find a counterexample that shows that $\sqrt{x}$ isn't the inverse of $2^x$: namely, take $x=64$. Then we have $$2^{\sqrt{x}}=2^8=256\not=x.$$
So it's clear that $\sqrt{x}$ can't be the inverse of $2^x$. But this isn't very satisfying, since it doesn't explain why.
To understand why $\sqrt{x}$ isn't the inverse of $x$, think about what it means: $\sqrt{x}$ is "the thing we need to square to get $x$." So $\sqrt{x}$ is the inverse of $x^2$: (the thing we need to square to get $x$) squared is obviously just $x$ itself.
I'm being a bit glib about square roots here - there's actually a serious issue, namely that the function $x^2$ isn't bijective and so doesn't have a genuine inverse. Basically, the function $\sqrt{x}$ isn't really "the thing we need to square to get $x$," but rather "the nonnegative thing we need to square to get $x$." The function $-\sqrt{x}$ also has the same nice behavior that $(-\sqrt{x})^2=x$. Really, $\sqrt{x}$ and $-\sqrt{x}$ are right inverses of $x^2$. But this is a bit of an aside from the general picture here.
The moral of the story: look at how the functions you're thinking about are defined. These definitions will often tell you directly what is an (almost) inverse of what; it's not always what you'd expect, but it can be figured out by thinking about what the functions mean.
A: To find the inverse we follow the following procedure:
The inverse of the function $x^2$:
Let $y = x^2$ (we need to write $x$ in terms of $y$)
$$y = x^2 \implies x = \sqrt{y}$$
Finally we change the places of $x$ and $y$. Therefore the inverse of the function $y=x^2$ is $y = \sqrt{x}$.
The inverse of the function $2^x$:
Let $y = 2^x$ (we need to write $x$ in terms of $y$)
$$y = 2^x \implies \log y = x \log 2 \implies x = \frac{\log y}{\log 2} = \log_2 y$$
Finally we change the places of $x$ and $y$. Therefore the inverse of the function $y=2^x$ is $y = \log_2 x$.
A: One of the reasons to explain this is if I take 2^x=y and find its inverse function I have to switch the x and y's and solve for y.
                        2^x=y

                        2^y=x        I have switched the x's and y's

                       \log_2 x=y    I solved for y

As a result, we see that the inverse of 2^x is a logarithmic function
Whereas, for x^3 or x^2 you can simply do the same thing.
                       x^2=y

                       y^2=x       Again, I switch the x's and y's

                       y=\sqrtx    Again, I solve for y

This shows that the inverse function of 2^x is, indeed, a logarithmic function and the inverses of any power functions such as x^3=y is a radical function.  
