How to find x in this case $\frac{x}{2}-1=e^x$?

Question

$$\frac{x}{2}-1=e^x$$

Let say we can write above function as: $$\frac{x}{2(e^x-1)}=1$$

In Wolframalpha I get a solution $x\approx -1.59362$.

How can I find x?

Answer 1

There is no solution in terms of elementary functions. Furthemore, there is no real solutuon, as shown in the answer by @cansomeonehelpmeout. However, you can use the Lambert W-function to get complex solutions:

$${x \over 2} - 1 = e^x$$ $$x-2 = 2e^x = 2e^{x-2}e^2$$ $$(x-2)e^{2-x}=2e^2$$ $$(2-x)e^{2-x}=-2e^2$$ $$2-x=W(-2e^2)$$ $$x=2-W(-2e^2)$$

Actually, $W$ is multi-valued. Taking the principal branch, we get $W_0(-2e^2) \approx 1.6718+2.2169i$, leading to $x\approx 0.3281-2.2169i$.

Answer 2

There is no real solution since

$$e^x\geq 1+x$$

For $x> -4$ we have $e^x\geq 1+x>\frac{x}{2}-1$, so $e^x-\left (\frac{x}{2}-1\right)>0$. This means that they don't intersect for $x\in \left <-4,\infty\right >$.

When $x\leq -4$ we have $e^x>0$ and so $e^x-\left (\frac{x}{2}-1\right )>0$ for $x\in \left <-\infty,-4\right]$. Therefore they don't intersect, and no real solution exists.

Answer 3

$$ \frac{x}{2} - 1 = e^x $$ $$ \frac{x}{2} - 2 = e^x - 1 $$ $$ \frac{x-4}{2} = e^x -1 $$ $$ \frac{x-4}{2(e^x -1)} = 1 $$

Let say we can write above function as:

You can't do that. Because then $x-4 = x \implies -4 = 0$, which is nonsense.

---

EDIT:

Assuming you meant:

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

Then you can just plot the function $f(x) = \frac{x}{2(e^x+1)}$ and solve $f(x) = 1$

EDIT 2:

I'm sure you're making a mistake: