Finding $b$ such that $y=mx+c$ is tangential to $y=b^x$. I was thinking about the following problem:

Find $b$ such that $y=mx+c$ is tangential to $y=b^x$.

My work so far:
To satisfy the condition we need to find $x$ such that $\frac d{dx}\left[b^x\right]=\frac d{dx}[mx+c]$. Solving the derivatives, we get $\ln b\cdot b^x=m\Rightarrow b^x=\frac m{\ln b}\Rightarrow x=\log_b\left(\frac m{\ln b}\right)=\frac{\ln \left(\frac m{\ln b}\right)}{\ln b}=\frac{\ln m-\ln\ln b}{\ln b}$.
At the same time, we have $b^x=mx+c$. Using the substitution $u=mx+c$ and noting $mx=u-c\Rightarrow x=\frac1mu-\frac cm$, we have:
$$b^{\frac um-\frac cm}=u$$
$$b^{-\frac cm}=u\left(\frac1b\right)^\frac um$$
$$ue^{\ln\left(\frac1b\right)\frac um}=ue^{-\frac{\ln b}mu}=b^{-\frac cm}$$
$$-\frac{\ln b}mue^{\ln\left(\frac1b\right)\frac um}=ue^{-\frac{\ln b}mu}=-b^{-\frac cm}\frac{\ln b}m$$
Let $z=-\frac{\ln b}mu$
$$ze^z=-b^{-\frac cm}\frac{\ln b}m$$
$$z=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$-\frac{\ln b}mu=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$-\frac{\ln b}mu=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$u=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$mx+c=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$mx=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$x=-\frac 1{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac cm$$
Putting the two equations together, we have:
$$\frac{\ln m-\ln\ln b}{\ln b}=-\frac 1{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac cm$$
$$\ln m-\ln\ln b=-W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac{c\ln b}m$$
$$\ln m-\ln\ln b+\frac{c\ln b}m=-W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$\ln\ln b-\ln m-\frac{c\ln b}m=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)e^{\ln\ln b-\ln m-\frac{c\ln b}m}=-b^{-\frac cm}\frac{\ln b}m$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)e^{\ln\ln b-\ln m-\frac cm\ln b}=-b^{-\frac cm}\frac{\ln b}m$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)\frac{\ln b}mb^{-\frac cm}=-b^{-\frac cm}\frac{\ln b}m$$
$$\ln\ln b-\ln m-\frac{c\ln b}m=-1$$
$$\ln\ln b-\frac{c\ln b}m=\ln m-1$$
$$e^{\ln\ln b-\ln b\frac cm}=e^{\ln m-1}$$
$$\ln b\cdot b^{-\frac cm}=\frac me$$
I was not able to get any further than this. The equation is probably not solvable through elementary means but I would be interested in a closed solution using special functions (such as the Lambert W function) as well as a derivation.
 A: Any point  on $y=b^x$ can be $P(t,b^t)$
The equation of the tangent at $P$ will be $$\dfrac{y-b^t}{x-t}=b^t\ln b$$
$$\iff y=x(b^t\ln b)+b^t(1-t\ln b)$$
Comparing with $y=mx+c$
$$m=b^t\ln b, c=b^t(1-t\ln b)$$
$$\ln m= t\ln b\cdot \ln(\ln b)\implies t=?$$
$$\dfrac cm=\dfrac{1-t\ln b}{\ln b}\implies \ln b=?, b=?$$
A: $y=b^x;\;y'=b^x\log b;\;b>0$ $
Derivative must be equal to the slope of $t:y=mx+c$
$$b^x\log b=m\to b^x=\frac{m}{\log b}\to x^*=\frac{\log \left(\frac{m}{\log (b)}\right)}{\log (b)}$$
Plug this value in the equation of the tangent. We get
$$y=m\,\frac{\log \left(\frac{m}{\log (b)}\right)}{\log (b)}+c$$
This value must be equal to the value we get substituting $x^*$ in the given equation
$$y=b^{x^*}\to y=\frac{m}{\log b}$$
The two $y$-values must be equal so we get the equation
$$m\,\frac{\log \left(\frac{m}{\log (b)}\right)}{\log (b)}+c=\frac{m}{\log b}$$
$$c \log b-\log (\log b)+m \log m=m$$
Set $\log b=u$
$$cu-\log u=m-m\log m$$
$$u -\frac{1}{c}\log u =\frac{m-m\log m}{c}$$
$$\log b=-\frac{m W\left(-\frac{c}{e}\right)}{c}$$
$$b=\exp\left(-\frac{m W\left(-\frac{c}{e}\right)}{c}\right)$$
where $W(z)$ is Lambert function.

Edit
Example.
If we have $y=2x-1$  we get
$$b=\exp\left(-2 W\left(-\frac{1}{e}\right)\right)\approx 1.7453$$
This is shown in the graph below

$$
...
$$

