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

Question

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.

Answer 1

$y=b^x;\;y'=b^x\log b;\;b>0$ $

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

---

$$ ... $$

Answer 2

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 $y=mx+c$

$$m=b^t\ln b, c=b^t(1-t\ln b)$$

$$\ln m= t\ln b\cdot \ln(\ln b)\implies t=?$$

$$\dfrac cm=\dfrac{1-t\ln b}{\ln b}\implies \ln b=?, b=?$$