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?
 A: Not without using the Lambert $W$ function, which is really just a fancy reformulation of the problem, and not actually a solution.
A: Let $f(x)=x-\log_2(x)-1$, which is the difference of the logarithms of the two members.
The derivative is $$1-\frac1{\ln(2)x},$$ which has a single zero for $x=\dfrac1{\ln(2)}$.
As there is a single extremum, there cannot be more than two roots, which you found.

There is no closed formula for these roots, except using the Lambert function. Isolating the roots in monotonic intervals so that they can be refined by numerical methods can/should be considered a valid algebraic solution.
A: 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.
A: $$\begin{align}
f(x)&=2^x-2x \\
f'(a)&=\ln2\cdot e^a-2\\
a&=\ln(2\ln2)\\
f'(a)&=2\ln2-2\ln(2\ln2)=2\ln\frac{1}{\ln 2}<0
\end{align}$$
Therefore $f'$ has two roots.
So, $x=1$ and $x=2$ are only roots.
A: Despite noting that an answer has already been chosen, I will attempt to show how problems like these are solved with the Lambert W function.$$2^x=2x$$$$1=\frac{2^x}{2x}=\frac 1 {2x}2^x$$Power everything we have by $-1$:$$1=1^{-1}=(\frac 1{2x}2^x)^{-1}=2x*2^{-x}$$From here, we attempt to make the exponent and coefficient the same with base $e$.$$1=2xe^{-x\ln(2)}$$$$\frac 1 2=xe^{-x\ln(2)}$$$$-\frac 1 2 \ln(2)=-x\ln(2)e^{-x\ln(2)}$$Now we can apply the Lamber W function, which is:$$y=W(ye^y)$$In other words, $y=-x\ln(2)$.  This gives us:$$W[-\frac 1 2 \ln(2)]=W[-x\ln(2)e^{-x\ln(2)}]=-x\ln(2)$$Now divide by $-\ln(2)$:$$x=\frac{W[-\frac 1 2 \ln(2)]}{-\ln(2)}$$.
Which, amazingly, should have solutions $x=1$ and $x=2$.
This is closest to $algebraic$ that you can do when given functions, one an exponential and one a polynomial (yes, you can do it for polynomials and even rational functions) and told they are equal to each other.
Also, if you had come upon, say, $10^x=x$, you would note that, graphically, the two lines do not cross.
However, with the Lambert W function, you can find the solution regardless if the graphs touch because the Lambert W function can be extended to complex and imaginary numbers.
I will also not the above examples as possible methods to solving your question with the use calculus.
In a nutshell, they found that a function $f(x)$ that equals zero.  You get this $f(x)$ as follows:$$2^x=2x$$$$\log_2(2^x)=\log_2(2x)$$$$x=\log_2(2*x)=\log_2(2)+\log_2(x)=1+\log_2(x)$$$$x=1+\log_2(x)$$$$x-\log_2(x)-1=0=f(x)$$Then we attempt to find $f'(x)$, the function that is the slope of $f(x)$ for any value $x$.$$f'(x)=1-\frac 1 {x\ln(2)}$$There are methods to finding this function when you learn calculus, so I won't elaborate.
The reason we solved for $f'(x)$ is so that we can find where the slope is zero.  This is helpful, for imagine a circle.  It has a slope of zero at the top and bottom.  Now imagine a polynomial.  It has a slope of zero at the tops and bottoms of all its curves.  This can be helpful for finding things like your solution.
In particular, we are finding $f'(x)=0$.$$0=1-\frac 1 {x\ln(2)}$$$$1=\frac 1 {x\ln(2)}$$$$x=\frac 1 {\ln(2)}$$
Now, let's find $f''(x)$, the second derivative of $f(x)$, or the slope of $f'(x)$.  We can use this to determine if a graph bends up or down.$$f''(x)=\frac 1 {x^2\ln(2)}$$For all $x$, $f''(x)$ is positive.  This means it always bends upwards.
Also, we have proven that the graph has a slope of zero at $x=\frac 1 {\ln(2)}$.  This means that from this point on, the graph goes up.  It has a "U" shape.  The bottom of this "U" is at $x=\frac 1 {\ln(2)}$.
Also recall that we are trying to find $f(x)=0$.
So from all of this, we have the following:
1) $f(x)=0$ happens at the x-intercepts.
2) $f'(x)=0$ is the bottom of our "U" shaped graph.
3) $f''(x)>0$, meaning it will always bend up in a "U" shape.
From this, I deduce that I should probably find $f(x)_{bottom.of."U"}$.  I know this occurs at $x=\frac 1 {\ln(2)}$, so I simply plug that into my $f(x)$ formula.$$f(\frac 1 {\ln(2)})=\frac 1 {\ln(2)}-\log_2(\frac 1 {\ln(2)})-1$$If this is negative, then my $f(x)$ function will go up (in a "U" shape) and hit the x-axis.  Where it hits the x-axis is your solution, if and only if $f(\frac 1 {\ln(2)})<0$.  Otherwise the "U" will never touch the x-axis.
And using a calculator, we see that it is in fact negative.
Which means we just proved the solution's existence (Meaning the solution is not imaginary or complex).
So what now?  This probably hasn't helped you much, but now that we see this exists, I would recommend Linear Approximation.  Sadly, there is no algebraic method to solving your problem from here.  If you want to attempt Linear Approximation, you go right ahead, because it is too tedious for my answer.  If you don't understand it, try youtube or a different Linear Approximation method because some methods are easier than others.
P.S.
Also, as a bonus, you already know what $f(x)$, $f'(x)$, and $f''(x)$ are.  Just look for them in my post.  Most linear approximation methods require those formulas.  I believe my functions have also already met all requirements for you to start attempting to use Linear Approximation, e.g., $f(x)=0$.
Simply and with much pleasure,
Simple Art.
