Solve equation: $5^x = -2x + 7$ How to solve that equation: $$5^x = -2x + 7$$
I already have the answer $x=1$. Can anyone please explain to me?
 A: $5^x$ monotonically increases, $-2x+7$ monotonically decreases. So, your equation couldn't have more than one root. And you've found it. 
A: Since the function $5^x+2x-7$ is continuous and you can see that it is monotonic increasing, by Brouwer's fixed point theorem (along with continuity ) the function has a unique zero which happens to be at $x=1$.
A: $x \gt 1 \implies 5^{x} \gt 5^{1} = 5 = -2 \times 1+7 \gt -2x +7$ 
$x \lt 1 \implies 5^{x} \lt 5^{1} = 5 = -2 \times1+7\lt -2x +7$ 
Hence only solution $x=1$.
A: The equation can be solved in terms of the Lambert W function.
Writing $5^x$ as ${\rm e}^{x \ln 5}$ the equation becomes
\begin{eqnarray*}
{\rm e}^{x \ln 5} &=& -2x + 7 \\
\Rightarrow (-2x + 7) {\rm e}^{-x \ln 5} &=& 1 \\
\frac{\ln 5}{2} (-2x + 7) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2} \ln 5 - \frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5+ \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2}\ln 5} {\rm e}^{-\frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} \\
\Rightarrow (-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} 
\end{eqnarray*}
The last equation is now in the form for the defining equation for the Lambert W function, namely
$${\rm W}(x) {\rm e}^{{\rm W}(x)} = x$$
Solving for W one has
$$-x \ln 5 + \frac{7}{2} \ln 5 = {\rm W}_0 \left (\frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \right )$$
Note the principal branch ${\rm W}_0(x)$ for the Lambert W function is chosen since its argument is positive. 
Finally, solving for $x$ yields
\begin{equation}
x = \frac{7}{2} - \frac{1}{\ln 5} {\rm W}_0 \left (\frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \right )
\end{equation}
which is the solution we sought. 
Getting the above solution into a form more easily recognised we note that since
$${\rm e}^{\frac{7}{2} \ln 5} = {\rm e}^{(\frac{5}{2} + 1) \ln 5} = {\rm e}^{\frac{5}{2} \ln 5} {\rm e}^{\ln 5} = 5 {\rm e}^{\frac{5}{2} \ln 5}$$
the equation for $x$ can be re-written as
$$x = \frac{7}{2} - \frac{1}{\ln 5} {\rm W}_0 \left (\frac{5 \ln 5}{2} {\rm e}^{\frac{5 \ln 5}{2}} \right )$$
Now as the Lambert W function is the inverse of the function $y = x {\rm e}^x$, one has the following simplification
$${\rm W}_0 \left (\frac{5 \ln 5}{2} {\rm e}^{\frac{5 \ln 5}{2}} \right ) = \frac{5 \ln 5}{2}$$
from which we get
$$x = \frac{7}{2} - \frac{1}{\ln 5} \frac{5 \ln 5}{2} = \frac{7}{2} - \frac{5}{2} = 1$$
