Knowing that $\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x=e$
and, $\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^{bx}=e^{ab}$
and also, $\lim_{x\to \infty}\left[\left(1+\frac{a}{x}\right)^{bx}+c\right]=e^{ab}+c$
How do you solve any equation with the form: $$y=\left(1+\frac{a}{x}\right)^{bx}+c$$
Where $a$, $b$ and $c$ are real numbers.
I know how to determine the horizontal asymptote of an equation with this form given the second equation above, but I do not know how to find the roots of an equation with this form.
To further clarify what I am asking, here is an example question with arbitrary values of $a$, $b$ and $c$:
Solve for x: $$0=\left(1+\frac{0.05}{x}\right)^{4x}-1.2$$ Not all equations in this form will have real roots, but for those that have soloutions, how would you find them?
2026-04-17 18:04:18.1776449058
Solve an equation with the form $y=\left(1+\frac{a}{x}\right)^{bx}+c$
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Considering that you are looking for the zero(s) of equation $$f(x)=\left(1+\frac{a}{x}\right)^{bx}+c$$ let $z=1+\frac{a}{x}$ which makes to equation to be $$f(z)=z^{\frac{a b}{z-1}}+c$$ Sooner or later, you will learn that any equation which can write or rewrite $$A+B y+C\log(D+Ey)=0$$ has solutions (real or complex) which can express in terms of Lambert function.
The Wikipedia page would show you many examples of the manipulations to be done before arriving to the proper form.
@dxiv gave the general solution for your problem. In terms of $z$, the solution would be $$z=-\frac{a b }{\log (-c)}W(A)\qquad \qquad\left(A=-\frac{ (-c)^{-\frac{1}{a b}}}{a b}\, \log (-c)\right)$$ and real solutions would be obtained if $A \geq -\frac 1e$.
The Wikipedia page gives series expansions of $W(A)$ as well as iterative schemes (based on Newton or Halley methods) to compute it. On the Internet, you should be able to find plenty of codes for the computation of this function.
If, for any reason, you cannot use Lambert function, only numerical methods, such as Newton, could be used.