$n=8\log_2(n)$, forgot basic math I'm quite ashamed that I'm at a math-related course at the university and I'm stuck. I can't solve at all this equation:
$$n=8\log_2(n).$$ 
I have tried applying the log property so it becomes
$2^n = 2^{8n}$.
Besides that this didn't help me, I'm not even sure if I applied right the property that says
$x=\log_2(y) \Rightarrow 2^x=y$
Thanks in advance!
Best Regards,
 A: It should be $2^{n/8}=n \implies 2^n=n^8$. Which cannot be solved analytically. Then you may want to use numerical methods to solve it.
A: You made a mistake. After applying $2^\cdot$ to both sides you obtain $2^n = n^8$. There will be two positive roots (you don't need negative because $n$ have to be positive since it is an argument of the logarithm).
There are two positive roots due to the fact that on $[0,1.1]$ $2^n>n^8$ but the derivative of $n^8$ is greater - so at approx. $n_1 = 1.1$ there is the first intersection. On the other hand exponential function grows faster then any power function and there will be the second root $n_2 \approx 43.5$. After this root the exponential function $2^n$ will be greater than $n^8$ as well as its derivative will be greater than the derivative of the power function for all $n>n_2$ hence there will no be more roots. 
Edited: There is no integer solution for an equation $100n^2 = 2^n$ since the left-hand side will never be a power of $2$ - the same for the rationals. On the other hand you can always use LambertW function which is defined as a solution of such equations (as it was mentioned in the answer which is deleted now). 
Due to the same reasons there will be two roots which you can find numerically.
A: This is hard to solve because there are no elementary manipulations that will convert this into a form you can use. That is, solving for $n$: 
$$2^n = n^8$$
is not a problem that is reasonable to be solved in a pre-calc or calculus type class.
In fact, there is a symbolic solution to this, but it's sorta cheating. Suppose you have a function $W(x)$ such that
$$y = W(x) e^{W(x)}$$ 
(This is called the Lambert W Function )
then you can use $W$, with suitable algebraic manipulation, to help solve weird relations like yours (um...Arturo did this for you so thankfully I won't have to bother now). 
Then you can use numerical properties of $W$ to get actual values.
The cheating part is because you just assume there's a solution function of a certain form $W$ with its properties) and manipulate your original problem to reduce to using $W$ (as Arturo did). And then, because other people have done lots of numerical analysis on $W$, you can (using tables or a computer algebra system like Maple or Mathmatica) get a numerical answer.
But if all you care about is a symbolic answer, you can stop at Arturo's.
Which is all to say, you shouldn't be ashamed at all, the equation you have is not solvable with the tools you have at your disposal.
A: *

*No, you didn't apply it correctly. Since $8\log_2(n) = \log_2(n^8)$, what you have is equivalent to $n=\log_2(n^8)$, or $2^n = n^8$. Equivalently, by taking $8$th roots, $2^{n/8} = n$, which can also be obtained by writing the original equation as $\frac{n}{8} = \log_2(n)$. 

*The reason you are having trouble is that there is no elementary solution. You can write a solution using Lambert's W function. The $W$ function has the property that $W(z)=x$ if and only if $z=xe^x$. 
To use Lambert's $W$ function, we can proceed as follows: let $u=\frac{n}{8}$. Then:
$$\begin{align*}
n &= 2^{n/8}\\
8u &= 2^u\\
\frac{8u}{2^u} &= 1\\
1 &= 8ue^{-u\ln(2)}\\
\frac{1}{8} &= ue^{-u\ln(2)}\\
\frac{-\ln 2}{8} &= -u\ln(2)e^{-u\ln(2)}\\
W\left(-\frac{\ln 2}{8}\right) &= -u\ln(2)\\
u &= -\frac{1}{\ln(2)}W\left(-\frac{\ln 2}{8}\right)\\
\frac{n}{8} &= -\frac{1}{\ln(2)}W\left(-\frac{\ln 2}{8}\right)\\
n &= -\frac{8}{\ln 2}W\left(-\frac{\ln 2}{8}\right).
\end{align*}$$
Added. Note that Lambert's $W$ function on the reals is "double valued" on $(-\frac{1}{e},0)$. Since $-\frac{\ln 2}{8}$ lies in this interval, this actually gives you two values for $n$.
Remark. I certainly wouldn't call Lambert's W function "basic math"; so let's say you got a bit confused with the logarithm properties and then ran into a big brick wall (which most people running this direction run into the first time, before they are told about the secret passage to go under it), rather than saying you "forgot basic math"...
