Solve for $n$ in $2^n=8$ So, I was wondering if it is possible to solve for $n$ in $2^n=8$ (or any other question where $n$ is a power) using $9^{th}$ grade math. Please excuse my naïveté if this is extremely stupid/simple. 
Thanks so much in advance! –– come to think of it: Is it possible at all?
 A: Note that $8=2^3$. Compare it to $2^n$ and conclude that $n=3$.
A: The best way to solve problems like this is using logarithms. If you have an equation 
$$10 ^x = y$$
where $y$ is any positive number, and you wish to find $x$, then the value of $x$ will be (by definition) $\log y$ (this is what the $\log$ button on your calculator is for). 
For example, to solve 
$$10^x = 110$$
we calculate $$x=\log 110= 2.0413...$$
There is a simple way to solve any equation like this. If $a$ and $y$ are positive, then the solution of 
$$a^x = y$$
is 
$$x= \frac{\log y}{\log a}$$
So in your example, the answer would be $\frac{\log 8}{\log 2}=3$. 

The theory of logarithms isn't too complicated, and logarithms were actually used as a way of multiplying large numbers before the invention of calculators (high school students would be given books of logarithms to use!). Here is a good place to start learning about them. 
However, at your level, the best way will be to use trial and error. For example, to solve $10^x=110$, we know that $10^2=100$ and $10^3=1000$, so $ x$ will be between $2$ and $3$. Similarly, we can show that $2<x<2.1$ and so on.
A: I don't know what is included in 9-th grade math, but here is a try to explain that $ 2^n, n \in \mathbb N $ tends to infinity as $n$ does so.
Obviously $n=3$ is a solution, since $2^3=8$. Now, if you try larger values of $n$ then $ 2^n$ becames larger and larger, since every time you multiply more times $2$ with itself. So, it is impossible to reach again $8$.
Edit(after Mathmo123 comment): Probably I misread the question, and I understood why is it unique. I apologize. If the question is " How we can solve this equation, i.e how to see that $n=3$ is a solution then we can work out a few values of $n$ since $8$ is small comparable to what we get raising $2^n$ for large n$. So, you try,
$n=1$: then, $2^1=2$
$n=2$: then, $ 2^2=4$
$n=3$: then, $\boxed{2^3=8}$
$n=4$: then, $2^4 =16>8$
$n=5$: then, $2^5= 32 > 8$
and so on...larger and larger.
For general exponential equations, we use logarithms as pointed out in the answer of Mathmo123.
