Solutions to $e^x + x = 2$

Question

I am extremely new to mathematics, and I don't have much training except for the basics so please excuse my rather basic question.

The question itself:

If I have the relationship $e^x + x - 2 = 0$;

and $k$ is the number of solutions in $[0,1]$ and $n$ is the number of solutions not in $[0,1]$

what is $k$? what is $n$?

My Questions (sorry if they are many)

(1) What is the proper name for this type of equation. It doesn't seem to be a function, just a relationship of one variable to some constants.

(2) How do you tell how many solutions an equation like this has just from looking at it? It seems like if you have an equation like $x^2 + 2x + 5 = 0$; you have at least two solutions. But what about the $e^x$? How do you predict how many solutions there will be then?

(3) Once I re-arrange the equation to $e^x + x = 2$; what algebraic steps could I follow if I wanted to find the exact solution?

Answer 1

1. You are confused about the meaning of the word "function". A function is a rule that transforms numbers into other numbers. "Take the last digit and put it at the front" is a function. I'm not sure if there is a proper name for this type of equation.

2. As far as I know there is no general rule. In the case of polynomials (where you just have x, multiplication, and addition, ie $x^5+5x^2=0.2$, but not $1/x=2$ because that involves division), there are at most as many solutions as the degree (the highest power of x), not at least.

3. Unfortunately, algebra is not purely mechanical. This is a serious misconception school gives us about mathematics: that once you express a problem as an equation or symbolic formula, the rest is just mechanically applying rules until you get the solution. You will find as you progress that many problems require a dose of ingenuity and creativity to discover their solution, not just algorithmic application of rules and steps. Your idea to consider seperately the solutions in $[0, 1]$ and $\neg[0, 1]$ is an example of the kind of "non linear" thinking that is often required. In this specific case, I don't know off the top of my head if it's even possible to obtain the solution via mere algebra.

Answer 2

Here are some answers: 1) These are called transcendental equations. Ultimately the objective is to find a root(or many) i.e. a solution either real or complex to these equations.

2) There are a few clues you can get. For instance in the original problem, $f(x) = e^x + x -2$ is monotone strict increasing and $f(0) = -1 < 0$ and $f(1) = e - 1 > 0$. If you have heard of intermediate value theorem, this will tell me that there is one unique real root to this equation in $[0,1]$.

3) There are iterative procedures such as Newton Raphson's method to get the answer. But to get it in a closed form is not always guaranteed. At least not to my knowledge.

Hope this helps.

Answer 3

You can solve this equation graphically in a first step to see how many solutions exist. So the question becomes how many $x\in \mathbb{R}$ exist such that $e^{x}=2-x$.

So, you can see that this $x$ is unique and positive. So the question is $x\leq 1$ or $x>1$

You can find your answer is you study the variation of the function $e^{x}+x-2$.

Answer 4

1) I don't know of any particular name given to these equations involving different types of functions.

2) There is no set procedure to determine the number of solutions without numerical methods. In this particular case, note that the function $f(x)=e^x+x$ is monotonically increasing, as it's derivative is strictly positive; so if there is an $x_0$ such that $f(x_0)=2$, it will be unique.

Now noting that $f(0)=1<2$ and $f(1)=e+1>2$, by the Intermediate Value Property of Continuous functions, there exists at least one $x_0\in [0,1]$ such that $f(x_0)=2$.

From $f$ being increasing, there is exactly one solution to $f(x)=2$ in $[0,1]$, and none outside it, i.e. $k=1$, $n=0$.

3)Again, I do not know of any non-numerical, let alone algebraic, methods to get the exact solution to your equation. Perhaps someone else will be able to help you further.