What is the inverse of $2^x$? Note: This may not be correct mathematical term, so in case of confusion, I mean what division is to multiplication. If not, just poke me in the comments.
I was given this the other day:
$2^x=8$
Which I solved as follows:
$2^x=2^3 \implies x=3$
What did I actually do to solve the equation?
Does this apply for other numbers and is there a general rule? I.e. $n^x$
 A: $$x = \log_2 y \iff y = 2^x$$
The inverse is called the base two logarithm. In your case $2^x = 8 \iff x = \log_2 8 = \log_2 2^3 = 3$.
In general the inverse for $a^x$, where $a> 1$ is the base $a$ logarithm. So $y = a^x \iff x = \log_a y$. 
A: Those cases are easy to solve if you have something like the example you just gave. Look for how does logarithm and exponential functions work.
In other cases like $2^x=\pi$ that is more complicated to computate manually. By computate manually I mean by giving an explicit value in decimal notation. So look some info about exponential and logarithmic functions, there is a lot of material about that in the internet.
A: The property you employed to conclude $x=3$ from $2^x = 2^3$ is called the injective property of the function $f(x)=2^x$.
A function is said to be injective (or 1-1) if $$f(a)=f(b) \implies a=b.$$
In the case of $f(x) = n^x$ for a fixed positive $n$ (except for $n=1$), the function is injective, so you can make the same conclusion. Thus $3^x = 27 \implies x=3$, $2^{x} = 1024 \implies x=10$ etc.
Functions that are injective, are invertible. This means that there is another function, $g$, whose domain is the range of the original function, $f$, for which $$g(f(x)) = x$$ for all $x$ in the domain of $f$.
The inverse function for the function $f(x) = n^x$ is called the logarithm of base $n$, $log_n(x)$. Note that $n$ does not have to be an integer, and a special base $e$ is associated with the natural logarithm: $\ln(x) = \log_e(x)$.
Thus $\log_2(2^x) = x$. We can use this to "solve" for $x$ in $2^x=8$:
$$2^x = 8$$
$$\log_2(2^x) = \log_2(8)$$
$$x = \log_2(2^3) = 3.$$
However, this is no different from using the injective property of $2^x$.
A: As the exponential function is strictly growing, the equation $2^x=y$ has a single real solution in $x$ (for $y>0$), which is called the logarithm (in base $2$), denoted as $\log_2(y)$.
$$2^{\log_2(y)}=y.$$
The values of $2^x$ for increasing integers $x$ are $1,2,4,8,16,32\cdots$, and for negative integers, $\dfrac12,\dfrac14,\dfrac18\dfrac1{16}\cdots$, so for instance, $\log_2(8)=3$.
The logarithm can also be computed for other values of $y$. Consider $2^x=5$, and look at the first powers of $5$,
$$5,25,125,625,3125\cdots$$
As $5^3=125\approx128=2^7$, we have $5\approx2^{7/3}$ and
$$\log_2(5)\approx\frac73=2.333333\cdots$$
As you can check,
$$2^{7/3}=\sqrt[3]{2^7}=5.0396842\cdots\approx5.$$
By using larger powers, you could find that the precise value is
$$\log_2(5)=2.3219280\cdots$$
It is worth to note that the logarithms in some base can be used for problems in another. As
$$y=2^{\log_2(y)}=5^{\log_5(y)}=(2^{\log_2(5)})^{\log_5(y)}=2^{\log_2(5)\log_5(y)},$$
$$\log_5(y)=\frac{\log_2(y)}{\log_2(5)}.$$
The logarithms are precomputed in printed tables, or evaluated by electronic calculators. They are commonly computed for base $10$, denoted $\log_{10}$ or simply $\log$, and a strange based denoted as $e=2.718281828\cdots$, denoted $\log_e$ or $\ln$. But that's another story.
