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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?

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    What do you mean by "solve"? Numerically? – 2017-01-14
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    @Jack Yes, basically I want to find the value of x that satisfies the equation – 2017-01-14
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    I'm not sure what the limits have to do with the question. – 2017-01-14
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    @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 – 2017-01-14
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    Are you looking for a closed form of the solution(s), or are you looking to determine whether solutions exist for some given $a,b,c$? – 2017-01-15
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    @dxiv Well I'm not sure if a closed-form expression exists for all equations of this type; but basically, I want to find $x$ for some given $a,b,c$ – 2017-01-15
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    The [solution](http://www.wolframalpha.com/input/?i=solve+(1+%2B+a+%2F+x)%5E(b+x)+-+c+%3D+0+for+x) in general involves the [Lambert $W$ function](https://en.wikipedia.org/wiki/Lambert_W_function). But the example you gave at the end has the simple solution $x = 1/4$ so maybe there is more context to the question, which you haven't shared. – 2017-01-15
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    @dxiv Yes, $x=1/4$ is the solution, but I don't quite understand the Lambert W function part. How do I substitute the values inside the Lambert W function? – 2017-01-15
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    @MattMath [Here](http://www.wolframalpha.com/input/?i=solve+(1+%2B+a+%2F+x)%5E(b+x)+-+c+%3D+0+for+x) is the general solution, and [here](http://www.wolframalpha.com/input/?i=solve+(1+%2B+a+%2F+x)%5E(b+x)+-+c+%3D+0+for+x+where+a%3D0.05,+b%3D4,+c%3D1.2) 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. – 2017-01-15
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    @dxiv Thanks for the help! Are there any other ways in solving this equation without using $W(x)$? – 2017-01-15
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    @MattMath Not in general, though some particular cases may have elementary solutions. – 2017-01-15
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    @dxiv Okay thank you! – 2017-01-15

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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.