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Why is $$\frac{e^{2x}-1}{e^x-1}$$ equal to $e^x+1$ ?

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    $\begingroup$ Because $(e^x - 1)(e^x + 1) = e^{2x} - 1$. $\endgroup$ – 3x89g2 Jun 29 '16 at 7:25
  • $\begingroup$ though beware $e^x=1$ i.e. $x=0$ $\endgroup$ – Henry Jun 29 '16 at 7:26
  • $\begingroup$ @GerryMyerson: yes I did (and thanks to your quick comment, I now do) $\endgroup$ – Henry Jun 29 '16 at 7:27
  • $\begingroup$ Consider rewriting with the substitution $t=e^x$. $\endgroup$ – Yves Daoust Jun 29 '16 at 7:40
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Substitute $t = e^x \implies e^{2x} = (e^x)^2 = t^2$, which translates the expression to $\dfrac{t^2 - 1}{t-1} = \dfrac{(t-1)(t+1)}{(t-1)}$. Then what is the result when replacing $t = e^x$?

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Note: $(e^x-1)(e^x+1)=(e^x)^2+e^x-e^x-1=e^{2x}−1$
Also note that the denominator $e^x-1\ne0\implies e^x\ne1\implies x\ne0$

$$\frac{e^{2x}−1}{e^x−1}=\frac{(e^x-1)(e^x+1)}{e^x−1}=e^x+1 \text{ where } x\ne0$$

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