Equation for slope of exponential function of arbitrary base

Question

Say I have some general exponential function defined as such;

$$f(x,a,b) = a\left(1-b^{-x}\right)$$

Given any arbitrary value for $a$, how can I solve for $b$ such that $\frac{d}{dx}f(0,a,b) = 1$? Is there a general equation for this?

Answer 1

You are solving for when $\frac{d}{dx}\,f(0,a,b)=1.$ Technically, you probably mean that you're plugging in $x=0$ after taking the derivative, and not before. The right way to write that expression is this: solve $$\left(\frac{d}{dx}\,f(x,a,b)\right)_{x=0}=1 $$ for $b,$ given $a$. So, moving forward, we have: \begin{align*} \frac{d}{dx}\,f(x,a,b)&=a\ln(b)\,b^{-x},\quad\text{so solve} \\ a\ln(b)&=1\\ \ln(b)&=\frac1a \\ b&=e^{1/a}. \end{align*}

Answer 2

Since the derivative of $a\left(1 - b^{-x}\right) = a\ln(b)b^{-x}$, the solution would be to solve the following equation:

$$a\ln(b)b^{-x} = 1$$

However, I hit a wall and got stuck trying to solve this function symbolically, so I gave in and punched the equation into WolframAlpha. This is the answer I was given back;

$$b=\sqrt[x]{-\frac{aW\left(-\frac{x}{a}\right)}{x}}$$

I don't doubt this is a correct answer, but what I don't understand is the use of the Lambert W-Function. Is this really necessary in the solution, or is WolframAlpha complicating it more than need be? In any case, it's use is inconvenient, as I don't have an easy way to approximate the function in my use case. As well, WolframAlpha wasn't able to tell me the steps it took to reach this point, so if someone could provide that for me, it'd be a great help.