For which value(s) of $b$, are $b\ln(x)$ and $b^{x} $tangent to each other?

Question

For which positive value(s) of $b$, are $b\ln(x)$ and $b^{x} $tangent to each other?

By setting the derivatives of the two curves equal to each other:

$$\dfrac{b}{x}=b^{x}\cdot \ln(b)$$ $$\dfrac{b}{x}-b^{x}\cdot \ln(b)=0$$

How to proceed from here?

Answer 1

Using a special function it is pretty easy to solve your equation, let me introduce you the Lambert W function defined by the following equation:

$$W(z)e^{W(z)}=z$$

Now just rewrite your equation as follows:

$$\frac bx=b^x\ln b\Rightarrow b=e^{x\ln b}x\ln b $$

Solving for $x$:

$$x\ln b=W(b)\Rightarrow x=\frac {W(b)}{\ln b}$$

EDIT: now if we want to solve for $b$ rearrange like:

$$\frac 1x=b^{x-1}\ln b$$

Then transform it this way:

$$\frac 1x=e^{(\ln b)(x-1)}\ln b\Rightarrow \frac {x-1}x=e^{(\ln b)(x-1)}(\ln b)(x-1)$$

That has solution:

$$(\ln b)(x-1)=W(\frac {x-1}x)\Rightarrow \ln b=\frac {W(\frac {x-1}x)}{x-1}\Rightarrow b=\exp(\frac {W(\frac {x-1}x)}{x-1})$$