# Question #ef937

Sep 4, 2017

${e}^{x - 1} + 1$

Could also write this as $\frac{{e}^{x} + e}{e}$ if that is more helpful.

#### Explanation:

$\frac{{e}^{x} \left(1 + {e}^{1 - x}\right)}{e} ^ 1$

Divide the terms outside the bracket (ie subtract indices)

$= {e}^{x - 1} \left(1 + {e}^{1 - x}\right)$

Now multiply through the bracket (ie add indices)

$= {e}^{x - 1} \cdot 1 + {e}^{x - 1} \cdot {e}^{1 - x}$

$= {e}^{x - 1} + {e}^{0} = {e}^{x - 1} + 1$

Sep 4, 2017

${e}^{x - 1} + 1.$

#### Explanation:

Assuming that, simplification for the Exression $\frac{{e}^{x} \left(1 + {e}^{1 - x}\right)}{e}$

is reqd.

$\text{The Exp.=} \frac{{e}^{x} \left(1 + {e}^{1 - x}\right)}{e}$

$= \left({e}^{x} / {e}^{1}\right) \left(1 + {e}^{1 - x}\right) ,$

$= {e}^{x - 1} \cdot \left(1 + {e}^{1 - x}\right) \ldots \ldots \ldots \ldots . \left[\because , {a}^{m} / {a}^{n} = {a}^{m - n}\right] ,$

$= {e}^{x - 1} \cdot 1 + {e}^{x - 1} \cdot {e}^{1 - x} \ldots \ldots . . \left[\because , l \left(m + n\right) = l m + \ln\right] ,$

$= {e}^{x - 1} + {e}^{\left(x -\right) + \left(1 - x\right)} \ldots \ldots \left[\because , {a}^{m} \cdot {a}^{n} = {a}^{m + n}\right] ,$

$= {e}^{x - 1} + {e}^{0} ,$

$\therefore \text{ The Exp.=} {e}^{x - 1} + 1 , \because , {a}^{0} = 1 , a \ne 0.$