I want to find out the values of x where the $f(x) = e^x-x-4$ will equal zero.
My problem by solving this myself is that I cannot use logarithm natural (ln) because I have a normal x:
$f(x) = e^x - x - 4$
$f(x) = 0 $
$\Rightarrow e^x - x -4 = 0 (| +x | +4)$
$\Rightarrow e^x = x + 4$
Failed solution one: $ e^x = x + 4 | ^{(-x)}$
$ e^0 = (x + 4)^{(-x)}$
$1 = (x + 4)^{(-x)}$ ?
Failed solution two:
$e^x = x + 4 | \ln$
$\Rightarrow x = \ln(x) + \ln(4)$ ?
What is the solution?
Cheers bodo
On
If you just need a numerical solution, then there are many methods available. For example, Newton's Method converges quite rapidly on each root. First, if $f(x) = e^x - x - 4$, then $f'(x) = e^x - 1$. Since $f'$ has only the root $x=0$, we can infer that $f$ has at most two roots, one on either side of $0$. Set up the recursive formula for Newton's Method: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{e^{x_n} - x_n - 4}{e^{x_n}-1}. $$ With initial guess $x_0 = -4$, and 4 iterations, we find a solution stable to 14 decimal digits: $-3.98133937091132$. With initial guess $x_0=2$, we get: $1.74903138601270$ (stable to 14 digits after 6 iterations).
This kind of equations generally have no closed-form solution.
It is often possible to manipulate the equation into a form that allows a solution to be written in terms of Lambert's W function, but that's usually not a major improvement from a practical point of view.
You will need to approximate the solution numerically.
Solution with $W$: Set $y = -e^x$. Then $x-e^x = -4$ becomes $$ \log(-y)+y = -4 $$ Exponentiating both sides get $$ -ye^y = e^{-4} $$ so by definition of the $W$ function we have $y=W(-e^{-4})$. Plug into the definition of $y$ and we get $$ x = \log(-W(-e^{-4}))$$ which produces one solution for each of the two branches of $W$. Enlightening, right?