Solve for $x$ using the lambert W function $ \frac{\ln(1+bx)}{x} = a$

Question

Question: Solve for $x$ using the lambert W function

<p>$$ \frac{\ln(1+bx)}{x} = a$$</p>

I've got this far:

$$ \frac{\ln(1+bx)}{x} = a$$

$$ \ln(1+bx) = ax $$

$$ 1+bx = e^{ax} $$

Stuck when to use the lambert W function

Answer 1

$$1 + bx = e^{ax}$$ Take $b$ common and multiply throughout by $e^{-ax}$ to get, $$\left(\frac{1}{b} + x\right)be^{-ax} = 1$$ Divide throughout by $b$ and multiply throughout by $e^{-\frac{a}{b}}$ to get, $$\left(\frac{1}{b} + x\right)e^{-ax}e^{-\frac{a}{b}} = \frac{1}{b}e^{-\frac{a}{b}} \Rightarrow \left(\frac{1}{b} + x\right)e^{-a\left(x + \frac{1}{b}\right)} = \frac{1}{b}e^{-\frac{a}{b}}$$ Multiply throughout by $-a$ to get, $$-a\left(x + \frac{1}{b}\right)e^{-a\left(x + \frac{1}{b}\right)} = -\frac{a}{b}e^{-\frac{a}{b}}$$ Note that the above is in the form $ye^y = c$, hence $y = W(c)$, $$-a\left(x + \frac{1}{b}\right) = W\left(-\frac{a}{b}e^{-\frac{a}{b}}\right)$$ Simplying for $x$ we get, $$x = -\frac{1}{a}W\left(-\frac{a}{b}e^{-\frac{a}{b}}\right) - \frac{1}{b}$$

Answer 2

Start with the equation

$$1+bx=e^{ax}$$

Then, let $y=1+bx$ so that

$$(-ay/b)e^{-ay/b}=(-a/b)e^{-a/b}$$

Therefore, we have

$$=W\left((-a/b)e^{-a/b}\right)=-ay/b$$

Solving for $y$, we find

$$y=-\frac ba W\left((-a/b)e^{-a/b}\right)$$

Finally, substituting back for $x$ yields

$$x=-\frac{1+\frac ba W\left((-a/b)e^{-a/b}\right)}{b}$$