What I'm stuck on is the way to get from $e^x-e^{x_0}$ to $x-x_0$. I know I could do a log base e to the $e^x-e^{x_0}$ but the algebra wouldn't result in $x-x_0$. I'm mostly just stuck on the logarithm and how to make $e^x-e^{x_0}$ turn into $x-x_0$.
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1Sometimes it is easier to use $x_0+h$ and $x_0$ rather than $x_0$ and a general $x$. For instance: $|e^{x_0+h}-e^{x_0}|=|e^{x_0}||e^h-1|$. – J. Moeller Oct 07 '18 at 02:59
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1this and this might help! – Chinnapparaj R Oct 07 '18 at 03:00
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@Prototank how can you use $x_0+h$ for x? – user2793618 Oct 07 '18 at 03:01
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@ChinnapparajR thanks for the links! – user2793618 Oct 07 '18 at 03:09