The equation $4^x=8x+1$ has only a solution. I need help to prove that the equation $4^x=8x+1$ has only one solution. I have only checked with some softwares that there is only a  solution  between $2$ and $3$ but I have problems with the formal proof.
Thanks in advance
Edit: Clearly $0$ is a solution, so I should prove that there is only a solution different from $0$.
 A: Hint : try to reason by absurd. Suppose there are 2 solutions ( $\neq 0$ ). Then consider the continuous and differentiable function $$f(x) = 4^x - (8x+1)$$
and denote as  $x_1$ and $x_2$ ($x_1$ and $x_2 \in [2;3]$) the two solutions of the equation $4^x=8x+1$. You can apply Rolle's Theorem with regard to $f(x)$ in the interval $]x_1;x_2[$ and obtain that there is a point $c$ in $]x_1;x_2[$ such that $f'(c)=0$. 
But $$f'(x)=D[4^x - (8x+1)]=4^x\ln(4)-8=0 \Rightarrow x=\log_4\left(\frac{8}{\ln4}\right)\approx 1.26$$ which is not in $[2;3] \to $ absurd conclusion, so there is only 1 unique solution in $[2;3]$.
A: Let $f(x)=4^x-8x-1$ and so
$$f(2)<0, f(3)>0$$
The derivative of $f$ says that $f$ has only a one critical point between 0 and 3 so the function attains minimum once i.e. the function vanishes at only one point between 0 and 3
A: Graph of function $f(x) =4^x - 8x + 1$ gives  you idea how to go about the proof. You show that $f$ is decreasing in $(-\infty,a)$ and increasing in $(a,\infty)$ and $f(a)<0, f(-\infty)=+\infty, f(\infty)=+\infty$. Then you know by monoticity that there is just one root in $(-\infty,a)$ and one in $(a,\infty)$.
A: Both answers can be explicitly found in closed formed in terms of the special function known as the Lambert W function.
To do this your equation needs to be first transformed into the form for the defining equation for the Lambert W function ${\rm W}(x)$ which is implicitly defined by the equation
$${\rm W}(x) {\rm e}^{{\rm \small{W}}(x)} = x$$
This can be done as follows:
\begin{eqnarray*}
4^x &=& 8x + 1 \\
{\rm e}^{x \ln 4} &=& 8x + 1 \\
(8x + 1) {\rm e}^{-x \ln 4} &=& 1 \\
\left (-x - \frac{1}{8} \right) {\rm e}^{-x \ln 4} &=& -\frac{1}{8} \\
\left (-x \ln 4 -\frac{\ln 4}{8} \right ) {\rm e}^{-x \ln 4 - \frac{\ln 4}{8} + \frac{\ln 4}{8}} &=& -\frac{\ln 4}{8} \\
\left (-x \ln 4 - \frac{\ln 4}{8} \right) {\rm e}^{-x \ln 4 - \frac{\ln 4}{8}} &=& - \frac{\ln 4}{8} {\rm e}^{-\frac{\ln 4}{8}} \\
\end{eqnarray*}
Which is now in the form of the defining equation for the Lambert W function. On solving one has
\begin{eqnarray*}
-x \ln 4 - \frac{\ln 4}{8} &=& {\rm W}_k \left (-\frac{\ln 4}{8} {\rm e}^{-\frac{\ln 4}{8}} \right ) \\
\end{eqnarray*}
As the argument for the Lambert W function lies between $-1/{\rm e} \leqslant x < 0$ both real branches $k$ of the Lambert W function need to be considered. For the two real branches one has $k = 0,-1$. 
Solving for $x$ gives
$$x = -\frac{1}{8} - \frac{1}{\ln 4} {\rm W}_k \left (-\frac{\ln 4}{8} {\rm e}^{-\frac{\ln 4}{8}} \right )$$
For the principal branch $k = 0$, one has the following simplification
$${\rm W}_0 \left (-\frac{\ln 4}{8} {\rm e}^{-\frac{\ln 4}{8}} \right ) = -\frac{\ln 4}{8}$$
Thus
$$x = \left\{
     \begin{array}{lr}
       0 \\
       -\frac{1}{8} - \frac{1}{\ln 4} {\rm W}_{-1} \left (-\frac{\ln 4}{8} {\rm e}^{-\frac{\ln 4}{8}} \right )
     \end{array}
   \right.$$
Numerically the second non-zero solution is equal to $x = 2.065 \, 693\ldots$
