Monique Slaughter

2021-12-13

How do you simplify ${e}^{-\mathrm{ln}x}$

zesponderyd

Use it for solution:
Consider the generic case of ${\mathrm{log}}_{10}\left(a\right)=b$
Another way of writing this is ${10}^{b}=a$
Suppose $a=10⇒{\mathrm{log}}_{10}\left(10\right)=b$
$⇒{10}^{b}=10⇒b=1$
So ${\mathrm{log}}_{a}\left(a\right)=1$ - important example
use this principle

Serita Dewitt

Write ${e}^{-\mathrm{ln}x}$ as $\frac{1}{{e}^{\mathrm{ln}x}}$
Let $y={e}^{\mathrm{ln}x}⇒\frac{1}{y}=\frac{1}{{e}^{\mathrm{ln}x}}$...Equation (1)
Consider just the denominators and take logs of both sides
$y={e}^{\mathrm{ln}x}\to \mathrm{ln}\left(y\right)=\mathrm{ln}\left({e}^{\mathrm{ln}\left(x\right)}\right)$
But for generic case $\mathrm{ln}\left({s}^{t}\right)\to t\mathrm{ln}\left(s\right)$
$⇒\mathrm{ln}\left(y\right)=\mathrm{ln}\left(x\right)\mathrm{ln}\left(e\right)$
But ${\mathrm{log}}_{e}\left(e\right)\to \mathrm{ln}\left(e\right)=1$
$⇒\mathrm{ln}\left(y\right)=\mathrm{ln}\left(x\right)×1$
Thus $y=x$
So Equation(1) becomes
$\frac{1}{y}=\frac{1}{{e}^{\mathrm{ln}\left(x\right)}}=\frac{1}{x}$
Thus ${e}^{-\mathrm{ln}x}=\frac{1}{x}$

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