How to solve for $c+ax=b^x$?

Question

So I am troubling myself on how to analytically solve for $x$ in $\text{c}+\text{a}x=\text{b}^x$.

Answer 1

The equation $y=xe^x$ has no inverse function that may be expressed using compositions of elementary functions (rational functions, radical functions, exponential and logarithmic functions, triginometric functions). Thus, just as we invented logarithms to invert exponentials, we invented a function to invert $f(x)=x e^x$. It is called the Lambert W function $W(x)$.

We generally don't invent new functions unless we need them. For instance, at least for positive numbers, we declared the inverse function of $x^2$ to be $\sqrt{x}$, but after that point we don't declare new inverse functions for every quadratic function $ax^2+bx+c$ (on an appropriate domain), because the $\sqrt{x}$ function is enough to write down those inverse functions as long as we mix it with other operations (as seen in the quadratic formula).

The equation $ax+c=b^x$ can be rearranged to be soluble by the Lambert W function as long as we introduce some substitutions along the way. To begin with,

$$\begin{array}{ll} ax+c=b^x & \iff (ax+c)b^{-x}=1 \\ & \iff (ax+c)\exp\big((-\ln b)x\big)=1. \end{array} $$

Set $u=(-\ln b)x$, in which case this becomes

$$\begin{array}{l} \iff \displaystyle\left(-\frac{a}{\ln b}u+c\right)\exp(u)=1 \\ \iff \displaystyle\left(u-\frac{c\ln b}{a}\right)\exp(u)=-\frac{\ln b}{a}. \end{array} $$

Finally, set $v=u-c\ln(b)/a$ to obtain

$$ \begin{array}{l} \displaystyle \iff v\exp\left(v+\frac{c\ln b}{a}\right) = -\frac{\ln b}{a} \\ \displaystyle \iff v\exp(v)= -\frac{\ln b}{a}\exp\left(-\frac{c\ln b}{a}\right) \\ \iff \displaystyle v=W\left(-\frac{\ln b}{a}\exp\left(-\frac{c\ln b}{a}\right)\right) \end{array} $$

Now, $\displaystyle v=u-\frac{c\ln b}{a}=(-\ln b)x-\frac{c\ln b}{a}$, so this becomes

$$ \begin{array}{l} \displaystyle \iff (-\ln b)x-\frac{c\ln b}{a}= W\left(-\frac{\ln b}{a}\exp\left(-\frac{c\ln b}{a}\right)\right) \\ \displaystyle \iff x=-\frac{c}{a}-\frac{1}{\ln b} W\left(-\frac{\ln b}{a}\exp\left(-\frac{c\ln b}{a}\right)\right). \end{array}$$

Answer 2

You can reduce it to $\enspace y=z^{\frac{1}{z}}$.

It exists several methods to solve this (e.g. Lambert W-function, different types of iteration, ...).

For your equation it's $\enspace \displaystyle y=(b^{1/a})^{b^{-c/a}}$ and $\enspace z=b^{c/a}(ax+c)$ .

$y$ is given, you solve first for $z$ and then for $x$.

The nice situation is here, that for $y>0$ you can immediately discuss the number of solutions without derivation:

$0<y\leq 1$ : one solution

$1<y<e^{\frac{1}{e}}$ : two solutions

$y=e^{\frac{1}{e}}$ : one solution $\enspace\enspace\enspace ($ here $ax+c$ is a tangent to $b^x$ $)$

$y>e^{\frac{1}{e}}$ : no solution