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