What is e^{-n(1/β-1/x)x} ?

The solution to e^{-n(1/β-1/x)x} is 1/(e^{\frac{n(x-β)){β}}}

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simplify e^{-n(1/β-1/x)x}

simplify

e^{- n (\frac{1}{β} - \frac{1}{x}) x}

\frac{1}{e ^{\frac{n ( x - β )}{β}}}

Solution steps

e^{- n (\frac{1}{β} - \frac{1}{x}) x}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{n x (- \frac{1}{x} + \frac{1}{β})}}

e^{n (\frac{1}{β} - \frac{1}{x}) x} = e^{n \frac{x - β}{β x} x}

= \frac{1}{e ^{n \frac{x - β}{β x} x}}

e^{n \frac{x - β}{β x} x} = e^{\frac{n ( x - β )}{β}}

= \frac{1}{e ^{\frac{( x - β ) n}{β}}}

e^{- (l n^{2} (x))}

9.8 \cdot (60.11 \cdot 1 0^{- 3}) \cdot 0.105

(2.3 \cdot 1 0^{- 3}) \cdot (6.02 \cdot 1 0^{23})

4.7 \cdot 1 0^{- 12}

\frac{1}{2} e^{- 2} \frac{3}{2} e^{- 2}

What is e^{-n(1/β-1/x)x} ?
The solution to e^{-n(1/β-1/x)x} is 1/(e^{\frac{n(x-β)){β}}}