What is e^{-(-x+1-ln(x-1))} ?

The solution to e^{-(-x+1-ln(x-1))} is (x-1)/(e^{-x+1)}

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simplify e^{-(-x+1-ln(x-1))}

simplify

e^{- (- x + 1 - l n (x - 1))}

\frac{x - 1}{e ^{- x + 1}}

Solution steps

e^{- (- x + 1 - l n (x - 1))}

Apply exponent rule:

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

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

Remove parentheses:

(- a) = - a

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

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

= \frac{1}{\frac{e ^{- x + 1}}{x - 1}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{x - 1}{e ^{- x + 1}}

6.912 (1 0^{- 4}) \cdot 12

2 π x^{6.671} x 1 0^{3}

\frac{Q ^{2}}{( b y + k y ^{2} ) ^{2}}

What is e^{-(-x+1-ln(x-1))} ?
The solution to e^{-(-x+1-ln(x-1))} is (x-1)/(e^{-x+1)}