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

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

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

simplify

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

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

Solution steps

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

Apply exponent rule:

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

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

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

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

(1 - e^{- x})^{- 1}

9.54 \cdot 1 0^{- 3}

e \cdot e^{2}

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