What is 1/((n-1)!)t^{n-1}e^{-at} ?

The solution to 1/((n-1)!)t^{n-1}e^{-at} is (t^{n-1})/(e^{ta)(n-1)!}

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simplify 1/((n-1)!)t^{n-1}e^{-at}

simplify

\frac{1}{( n - 1 ) !} t^{n - 1} e^{- a t}

\frac{t ^{n - 1}}{e ^{t a} ( n - 1 ) !}

Solution steps

\frac{1}{( n - 1 ) !} t^{n - 1} e^{- a t}

Apply exponent rule:

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

= \frac{1}{( n - 1 ) !} \cdot \frac{1}{e ^{t a}} t^{n - 1}

Multiply fractions:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

= \frac{1 \cdot 1 \cdot t ^{n - 1}}{( n - 1 ) ! e ^{a t}}

Multiply:

1 \cdot 1 \cdot t^{n - 1} = t^{n - 1}

= \frac{t ^{n - 1}}{e ^{t a} ( n - 1 ) !}

x^{\frac{2}{3}} y^{\frac{2}{3}}

e^{- 0.37 \cdot 222}

\frac{e ^{1}}{e ^{0.5}}

2.03688 x 1 0^{- 4}

2 a^{2} 0.3 b^{2}

What is 1/((n-1)!)t^{n-1}e^{-at} ?
The solution to 1/((n-1)!)t^{n-1}e^{-at} is (t^{n-1})/(e^{ta)(n-1)!}