What is expand-e^{an}(e^{-xn}-e^{-an}) ?

The solution to expand-e^{an}(e^{-xn}-e^{-an}) is -e^{an-nx}+1

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expand-e^{an}(e^{-xn}-e^{-an})

expand

- e^{an} (e^{- x n} - e^{- an})

- e^{an - n x} + 1

Solution steps

- e^{an} (e^{- x n} - e^{- an})

= - e^{an} (e^{- n x} - e^{- an})

Apply the distributive law:

a (b - c) = ab - a c

a = - e^{an}, b = e^{- x n}, c = e^{- an}

= - e^{an} e^{- x n} - (- e^{an}) e^{- an}

Apply minus-plus rules

- (- a) = a

= - e^{an} e^{- n x} + e^{an} e^{- an}

Simplify

- e^{an} e^{- n x} + e^{an} e^{- an} : - e^{an - n x} + 1

= - e^{an - n x} + 1

\frac{s + 2}{( s + 3 ) ( s ^{2} + 2 s + 2 )}

a \cdot (b + c + d + e) + b \cdot (c + d + e) + c \cdot (d + e) + d \cdot e

1 + (\frac{( e ^{x} - e ^{(- x)} )}{( 2 )})^{2}

\frac{x + 5}{( x + 2 ) ^{2}}

4 (\frac{x}{3} + 6 = 40)

What is expand-e^{an}(e^{-xn}-e^{-an}) ?
The solution to expand-e^{an}(e^{-xn}-e^{-an}) is -e^{an-nx}+1