# Exponent Manipulation

I'm noob in Mathematics. Currently I'm experiencing with Exponent Equation. I know exponent can be added if we multiply and subtract if we divided. But I'm lost how the following first equation sorts out to the second one.

$$a = \frac{e^3}{e^3 + e^2}$$ $$a = \frac{1}{1 + e^{2-3}}$$

e is the Euler's number (base of the natural logarithm). But the equation works with any base number.

• Divide numerator and denominator by $e^3$ – saulspatz Jul 3 '18 at 6:45

$$\frac{e^3}{e^3 + e^2}=\frac{\color{red}{\frac1{e^3}}\cdot e^3}{\color{red}{\frac1{e^3}}(e^3 + e^2)}=\frac1{1+e^{2-3}}=\frac1{1+\frac1e}$$

Use $$\frac x{y+z}=\frac x{x\cdot(\frac yx+\frac zx)}=\frac1{\frac yx+\frac zx}.$$

Hint:

Use $2=2-3+3$, hence $e^2=e^{2-3+3}=e^{2-3}e^3$.

$$a = \frac{e^3}{e^3 + e^2}$$ -- (1)

Factor out $$e^{3}$$ from the denominator in (1). That is, $$a$$ $$= \frac{e^3}{e^{3}(e^{3 - 3} + e^{2 - 3})}$$ $$= \frac{e^3}{e^{3}(e^{0} + e^{- 1})}$$ -- (2)

From (2), we conclude via this rule: $$x^{0} = 1$$ where $$x ≠ 0$$ that $$a = \frac{1}{1 + e^{- 1}}$$