I understand how my teacher got the two $x$ values, but why didn't he solve for $e^x=0$?
I know he did
$x=0$ which is $0$
$x+2=0$ which is $-2$
so why no $e^x=0$? is there even an answer for that? I don't think there is right?
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2$e^x\neq 0$ for all $x$, in fact $e^x>0$ if $x$ is real. – Suzu Hirose Nov 12 '14 at 01:49
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3Graph $e^x$ and see for yourself. – dustin Nov 12 '14 at 01:51
1 Answers
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If you look closely at the definition of $e^x$ you can conclude that there is no value such that $e^x = 0$ as seen in the formula below
$$ e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!} $$ Since when k = 0, the sum will be:
$$ e^x = 1+ \sum_{k=1}^{\infty}\frac{x^k}{k!} $$ Since all terms when summed together will yield a result greater than zero, there is no possible way to have any real value for $x$ such that $e^x = 0$. This is even true for any value you make x, including imaginary values of x.
Eric L
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@amzoti yes I edited my answer, I'm so used to putting n, quite a bad habit. – Eric L Nov 12 '14 at 02:18
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All other terms (other than the constant term) don't have to sum to something positive, but they do have to something greater than -1. – bzc Nov 12 '14 at 02:35