How to solve $2^x=2x$ analytically. $$2^x=2x$$
I am able to find the solutions for this equation by looking at a graph and guessing. I found them to be $x_1=1$ and $x_2=2$. 
I also can also find them by guess and check, but is there anyway algebraically to solve this problem?

Well there exists analytical way. Here is my method.
Firstly divide by 2 both sides to obtain $2^{x-1}=x$. Now let's construct the function : $f(x)=2^{x-1}-x$. The first derivative of which is $f'(x)=ln2*2^{x-1}-1$. And the second derivative of which is $f"(x)=(ln2)^2*2^{x-1}$. 
Now we can easily see that for all $x \rightarrow f"(x)>0$. So it follows that $f'(x)$ is increasing for all $x$. If for $x=2$ $f'(x)>0$, then for all $x>2$ $f(x)>0$. 
And that means for $x\ge 2$ the function $f(x)$ is increasing. Since $f(2)=0$ that means there is no root greater than 2.
In the same way we have that for $x\le 1$, $f(x)$ is decreasing. since $f(1)=0$ that means there is no root smaller than 1.
Also the rest is easy to show that between 1 and 2 there is no root.
