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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.

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  • $\begingroup$ Divide numerator and denominator by $e^3$ $\endgroup$ – saulspatz Jul 3 '18 at 6:45
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$$\frac{e^3}{e^3 + e^2}=\frac{\color{red}{\frac1{e^3}}\cdot e^3}{\color{red}{\frac1{e^3}}(e^3 + e^2)}=\frac1{1+e^{2-3}}=\frac1{1+\frac1e}$$

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

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Hint:

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

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

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