How to solve this non-algebraic equation: $e^x-5x+7=0$

Question

So I was bored, and decided to start solving equations that weren't algebraically solvable. After a while, I came up with this seemingly simple equation$$e^x-5x+7=0$$that I thought that I might be able to solve. Here is my attempt at doing this:$$e^x-5x+7=0\\\implies e^x=5x-7\\\implies1=(5x-7)e^{-x}\\\implies-\dfrac15=\left(-x+7/5\right)e^{-x}\\\implies-\dfrac{e^{7/5}}5=(-x+7/5)e^{-x+7/5}\\\implies W(-e^{7/5}/5)=-x+7/5\\x=7/5-W(-e^{7/5}/5)\\\approx0.2(7-5W_n(-0.81104)),n\in\mathbb Z$$according to Wolfram Alpha with the only approximate form I could get, which confirms that I am correct. However, why are we getting$$0.2(7-5W_{\color{red}{n}}(-0.81104)),n\in\mathbb Z$$for our approximation instead of just$$0.2(7-5W(-0.81104))$$

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Why is this?

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Well, this is because the product log is what is known as a multivalued function.

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What does this mean?

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Well, in math, when we say that a function is multivalued, we mean that a function is a set valued function that contains additional properties based on context. For example, take $\log(x)$ or $\ln(x)$. We know that the log base n of $x$ is how many times our base $n$ multiplies into our number $x$. Now, why is this considered multivalued when it's clearly a one-to-one function for $x\gt0$? Well, let's look at $\ln(-x),x\gt0$ (since the logarithm functions are all undefined for $\log_n(0)$, and for when we have $n=0$) We have$$\ln(-x)=\ln(-1)+\ln(x)\implies i\pi\color{red}{+2i\pi k}+\ln(x)$$because we have for any $-1=e^{z}$ we have $z=i\pi$, and this equation will remain true for every multiple of $2i\pi k,k\in\mathbb Z$, although I think this has been proved lots and lots of times before, so it's best not to waste time proving it. Just know it has something to do with sine and cosine's period, since trigonometric functions are also periodic.

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Okay, but why is Lambert's function $W(x)$ a multivalued function?

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Since I feel like I've probably already bored you with the previous two parts, I'll try to keep this short. The general idea is (you can read more about it on the Wikipedia here), we have that $we^w=z$ (for $w,z\in\mathbb C$) holds iff $w=W_k(z)$ for some integer $k$. Now, since we have $w,z\in\mathbb C$, we have$$we^w=z\implies (x_1+y_1i)e^{(x_1+y_1i)}=(x_2+y_2i)\\\implies(x_1+y_1e^{i\pi/2\color{red}{+2i\pi k}})e^{x+y_1i}=(x_2+y_2e^{i\pi/2\color{red}{+2i\pi k}})$$and since we have a period (through our complex number $i=\sqrt{-1}$), this can only hold iff$$w=W_k(z)$$for some integer $k$, hence $W(x)$ is a multivalued function.

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My question

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Are there any easier ways that I could have gotten my solution to my non-algebraic equation, or is this the easiest way?

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Notes

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1. While I doubt that there is an easier way to solve this, I do know of another way that I could have solved this, although it would have been more difficult: fixed point iteration.
2. Also am unsure whether to add the complex-analysis tag

Answer 1

Why, you ask, is $W$ a multivalued function? By definition, $W(z)$ is supposed to be a solution of $w e^w = z$. It turns out that for any $z \ne 0$, this equation has infinitely many solutions. Each of those is a branch of $W(z)$.

Note that $f(w) = w e^w$ has an essential singularity at $\infty$ (it can't be a pole because $f(w) \to 0$ as $w \to -\infty$ on the real axis, and can't be a removable singularity because $f(w) \to +\infty$ as $w \to +\infty$ on the real axis). By the Great Picard Theorem, $f(w) = z$ must have infinitely many solutions for all $z$ with at most one exception. That exception, of course, is $0$, as $w e^w = 0$ only for $w=0$.

Answer 2

COMMENT (Just for fun and certainly in conexion with the multivalued functions $W_n$. This is not an answer at all but only a pertinent comment with the O.P.).

►There is not real solution because $e^x\gt5x-7$ so putting $z=x+yi$ instead of $x$ we have $e^{x+yi}-5(x+yi)+7=0$ which gives $$\cos(y)e^x-5x+7=0\\\sin(y)e^x-5y=0$$ whose graphs are partially show in the attached figure.