Solve equation: $5^x = -2x + 7$

Question

How to solve that equation: $$5^x = -2x + 7$$ I already have the answer $x=1$. Can anyone please explain to me?

Answer 1

$5^x$ monotonically increases, $-2x+7$ monotonically decreases. So, your equation couldn't have more than one root. And you've found it.

Answer 2

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 $x=1$.

Answer 3

$x \gt 1 \implies 5^{x} \gt 5^{1} = 5 = -2 \times 1+7 \gt -2x +7$
$x \lt 1 \implies 5^{x} \lt 5^{1} = 5 = -2 \times1+7\lt -2x +7$
Hence only solution $x=1$.

Answer 4

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 $$x = \frac{7}{2} - \frac{1}{\ln 5} \frac{5 \ln 5}{2} = \frac{7}{2} - \frac{5}{2} = 1$$