# Solve an equation with the form $y=\left(1+\frac{a}{x}\right)^{bx}+c$

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?

• What do you mean by "solve"? Numerically? – Jack Jan 14 '17 at 21:42
• @Jack Yes, basically I want to find the value of x that satisfies the equation – MattMath Jan 14 '17 at 21:48
• I'm not sure what the limits have to do with the question. – Karl Jan 14 '17 at 21:49
• @Karl The original equation with the form y=(1+(1/x))^x has a horizontal asymptote of e. I was thinking the limits might help in solving the equation in some way – MattMath Jan 14 '17 at 21:52
• @MattMath Here is the general solution, and here is the solution for that particular case. To derive the former then specialize it to the latter, both require some familiarity with $W(x)$ and the associated identities. – dxiv Jan 15 '17 at 3:41

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.
@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.