Unknown both as a exponent and as a term in an equation

Question

Let's say I have an equation $e^{x-1}(x+1)=2$. According to Solving an equation when the unknown is both a term and exponent it's impossible to solve this using elemetary functions. If so, then how do you solve it? Could you give me some keywords for further research on solving this type of equations? I'd be even more grateful if you solved the equation above (if it's not too hard).

Answer 1

One easy solution is $x = 1$ which you can find by inspection. $e^{1-1} = e^0 = 1$ but $1+1 = 2$. So $e^{1-1}(1+1) = 2$.

The solution(s) are not going to be solvable using elementary techniques. In fact the solution(s) are $$x = W_n(2e^2) - 1$$ where $W_n$ is the analytical continuation of the product log function. Note: $W_n(2e^2) -1 = 1$

Otherwise, the other methods are numerical ones such as Newton-Rhapson, interval bisection, etc... to find the roots.

Answer 2

Set, for simplicity, $x-1=t$, so the equation becomes $e^t(t+2)=2$. Consider the function $$ f(t)=e^t(t+2)-2 $$ which is defined for all real $t$. Next $$ \lim_{t\to-\infty}f(t)=-2, \qquad \lim_{t\to-\infty}f(t)=\infty $$ (prove it). So we see that at least one solution exist, because by the intermediate value term, $f$ must assume the value $0$.

Let's have a look at maxima and minima: $$ f'(t)=e^t(t+2)+e^t=e^t(t+3) $$ that's quite easy to study: $f'(t)<0$ for $t<-3$ and $f'(t)>0$ for $t>-3$. Thus $f$ is decreasing for $t<-3$ and increasing for $t>-3$.

Since $f(-3)=-e^{-3}-2<0$, we see that there is only one point $t_0$ where $f(t_0)=0$. Since obviously $f(0)=0$, we found it.

So the only solution of your equation is $x=1$.

Answer 3

The Lambert W function is defined as $z=W(z)e^{W(z)}$. We have

$$(x+1)e^{x-1}=2 \tag 1$$

Multiplying both side of $(1)$ by $e^2$ reveals that

$$\begin{align} (x+1)e^{x+1}&=2e^2\\\\ &\implies x+1=W(2e^2)\\\\ &\implies x=W(2e^2)-1\\\\ &\implies x=2-1\\\\ &\implies x=1 \end{align}$$

---

NOTE:

For the more general case for which

$$(ax+b)e^x=c \tag 2$$

we can easily obtain the solution for $x$ in $(2)$ as

$$\begin{align} (x+b/a)e^{x}&=c/a\\\\ &\implies (x+b/a)e^{x+b/a}=\frac{c}{a} e^{b/a}\\\\ &\implies x=W\left(\frac{c}{a} e^{b/a}\right)-(b/a)\\\\ \end{align}$$

$$\bbox[5px,border:2px solid #C0A000]{x=W\left(\frac{c}{a} e^{b/a}\right)-(b/a)}$$

For the aforementioned problem, $a=b=e^{-1}$ and $c=2$. Thus, $x=W(2e^2)-1=1$ as expected!!

Answer 4

Seems to be a case for the Lambert W-function: \begin{align} e^{x-1}(x+1) &=2 \iff \\ e^{x+1}(x+1) &=2e^2 \iff \\ f(x+1) &= 2e^2 \Rightarrow \\ x + 1 &= f^{-1}(2e^2) \Rightarrow \\ x &= f^{-1}(2e^2) - 1 \end{align} where $f(x) = x e^x$ and $f^{-1} = W$, which is that Lambert W-function.

We need the branch with positive real numbers, which is called $W_0$: $$ x = W_0(2e^2) + 1 $$ Using $W_0(x e^x) = W_0(f(x)) = x$ we see $W_0(2e^2) = 2$ and $$ x = 2 - 1 = 1 $$

Note: This particular problem only needed the existence of an inverse to $f(x) = x e^x$. We could have written $$ x = f^{-1}(2e^2) - 1 = f^{-1}(f(2)) - 1 = 2 - 1 = 1 $$ without knowing about $W_0$. The existence of an inverse could be argued by the monotonicity of $f$.