Originally Posted by

**gerardhoyle** And if that proves that $\displaystyle e^x = \frac{1}{e^x}$ then I could use this fact to prove:

$\displaystyle \frac{e^x}{e^y} = e^{x - y}$

by

$\displaystyle e^{x - y} = e^x \cdot e^{-y} = e^x \cdot \frac{1}{e^y} = \frac{e^x}{e^y}$

Then smushing it all together to prove:

$\displaystyle (e^{x_1})^{x_2} = e^{x_1x_2}$

by;

$\displaystyle y_1 = e^{x_1}$

$\displaystyle y_2 = y_1^{x_2}$

$\displaystyle lny_1 = x_1$

$\displaystyle lny_2 = x_2lny_1 = x_1x_2$

$\displaystyle lny_2 = x_1x_2$

$\displaystyle y_2 = e^{x_1x_2}$

(Nerd)

(Thanks, if there's no error here then thanks for pointing me in the right direction, darn book!).