In his 1684 text, "A New Method for Finding Minima and Maxima," Leibniz solves an inverse-tangent problem posed by De Beaune that basically asks what kind of curve will have a constant subtangent (p. 8 in this translation) (i.e. the subtangent will always have length a in the picture here).
I understand that this is the same as asking for the solution to the differential equation:
$\frac{dw}{dx} = \frac{w}{a}$
which is an exponential function of the family:
$w = Ce^{x/a}$
where C is the initial value. This is equivalent to $x = a \log(\frac{w}{C})$.
What I don't understand is how to connect this solution to the way Leibniz actually solves the problem. Leibniz says that:
that is if the x shall be in an arithmetic progression, the w shall be in a geometric progression, or if w shall be numbers, x will be their logarithms: therefore the line WW is logarithmic.
Leibniz shows that $x$ forms an arithmetic progression and $w$ forms a geometric progression with common ratio $(1 + \frac{b}{a})$:
$x, x + dx, x + 2dx,..., x + ndx$
For w, Leibniz lets $dx$ be a constant $b$
$\frac{a}{b}dw = {w}$
$w + dw = w + \frac{w}{a}dx = w + \frac{b}{a}w = w(1 + \frac{b}{a})$
We can write the $n$th term of this geometric progression as $W_n = W_0(1 + \frac{b}{a})^n$. If we let $W_0 = C$ and let $x = n$, can we obtain the equation $x = a \log(\frac{w}{C})$ from this equation for the geometric sequence itself?
You can't let $x=n$ because $n$ is supposed to be infinite, whereas $x$ is finite.
Assuming for simplicity that $x$ ranges through $[0,1]$, we obtain that $b=dx=\frac1n$. Therefore in your notation, $W_n = W_0 \big(1+ \frac{a^{-1}}{n}\big)^n\approx W_0 \, e^{1/a}$, where $\approx$ is the relation of infinite proximity. Note that Leibniz routinely dropped negligible terms in his calculations.