# Thread: Algebra problem with unknown value on both sides

1. ## Algebra problem with unknown value on both sides

Hi all,

I'd really appreciate it if some one could show me what steps are required to solve the following equation with the ^ symbol representing power. I get the feeling it should be very simple but I don't seem to be able to work it out:

256^n = n

I've tried to rearrange the equation using the and rules but don't seem to be getting any where using this approach.

Thanks in advance for any help any one can give me.
Neil

2. Originally Posted by nastev
Hi all,

I'd really appreciate it if some one could show me what steps are required to solve the following equation with the ^ symbol representing power. I get the feeling it should be very simple but I don't seem to be able to work it out:

256^n = n

I've tried to rearrange the equation using the and rules but don't seem to be getting any where using this approach.

Thanks in advance for any help any one can give me.
Neil
Isn't it true that y = n and b = 256? If you're talking about natural logarithms, the final step would be to apply e to both sides of the equation.

3. Thanks for the reply. I'm still not understanding how apply e will help. (though I may be missing something obvious since my algebra is pretty rusty)

If I apply the log formula I get n = log256(n) (where log256 means log base 256). I don't understand where I go from here to derive n though. Possibly I might be taking the wrong approach to solving this by trying to rearrange it with logs?

4. Originally Posted by nastev
Hi all,

I'd really appreciate it if some one could show me what steps are required to solve the following equation with the ^ symbol representing power. I get the feeling it should be very simple but I don't seem to be able to work it out:

256^n = n

I've tried to rearrange the equation using the and rules but don't seem to be getting any where using this approach.

Thanks in advance for any help any one can give me.
Neil
First of all you need only look at $n$ where $n > 0$, since $f(n) < 0$ isn't really important here.

Then look at the growth rates of $f(n) = n$ and $f(n) = 256^n$.

The latter grows at an astonishing rate, while the other grows gradually, moving equally along the x and y axis. Wherever $f(n) = n$ is, $f(n) = 256^n$ will always be much further above it on the graph.

5. Just logarithms aren't going to help you. Here's what I would do:

From $256^n= n$, divide both sides by $256^n$ to get $1= n\left(\frac{1}{256}\right)^n$. Since $a^b= e^{ln(a^b)}= e^{b ln(a)}$, we can write $\left(\frac{1}{256}\right)^n= e^{n (-ln(256))}$ so $n\left(\frac{1}{256}\right)^n= ne^{n(-ln(256))}= 1$

Now multiply both sides by -ln(256):
$\left(n(-ln(256))e^{n(-ln(256))}= -ln(256)$

Let x= n(-ln(256)) and that becomes xe^x= -ln(256).

That cannot be solved in terms of "elementary functions" but It can be solved using "Lambert's W function" (also called the "omega function" and the "productlog function") which is specifically defined as the inverse function to $xe^x$. Applying that to both sides of this equation,
W(xe^x)= x= W(-ln(256)).

Since x= n(-ln(256)), $n= -\frac{x}{ln(256)}$ so that $n= -\frac{W(-ln(256))}{ln(256)}$.

The W function can be expanded in the MacLaurin series
$\sum_{k=1}^\infty \frac{(-1)^{k-1}k^{k-2}}{(k-1)!}x^k$