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?

  • $\begingroup$ Note that $\frac{d}{dx}\,f(x,a,b)=a\ln(b)\,b^{-x}.$ Can you go from there? $\endgroup$ – Adrian Keister Jul 12 at 13:19
  • $\begingroup$ I've managed to simplify the equation down to $b^{b^{-x}}=e^{\frac{1}{a}}$, but I can't make it any further on my own. Do you have any advice? $\endgroup$ – Maurdekye Jul 12 at 13:54

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*}

  • $\begingroup$ You've left out the $b^{-x}$ in one step of your equation. $\endgroup$ – Maurdekye Jul 12 at 14:22
  • $\begingroup$ You specifically said to solve $f'(0,a,b)=0,$ where you've plugged in $x=0.$ When you do that, you get $b^{-0}=1.$ $\endgroup$ – Adrian Keister Jul 12 at 14:23
  • $\begingroup$ Hmm, you're right. This appears to be the correct answer. $\endgroup$ – Maurdekye Jul 12 at 14:25

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;


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.

  • $\begingroup$ The Lambert $W$ function is there because you still have $x$ in your expression. It's the inverse of the function $f(x)=xe^x,$ and is useful at times. $\endgroup$ – Adrian Keister Jul 12 at 14:28

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