What is (t-1)e^{-a(t-1)}u(t-1) ?

The solution to (t-1)e^{-a(t-1)}u(t-1) is e^{-a(t-1)}u(t-1)^2

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simplify (t-1)e^{-a(t-1)}u(t-1)

simplify

(t - 1) e^{- a (t - 1)} u (t - 1)

e^{- a (t - 1)} u (t - 1)^{2}

Solution steps

(t - 1) e^{- a (t - 1)} u (t - 1)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(t - 1) (t - 1) = (t - 1)^{1 + 1}

= e^{- a (t - 1)} u (t - 1)^{1 + 1}

Add the numbers:

1 + 1 = 2

= e^{- a (t - 1)} u (t - 1)^{2}

8^{- l o g_{8} (3)}

1 \cdot j (1^{x^{d}})^{0.1}

7.14 \cdot 1 0^{- 3}

(3) (1.6 \cdot 1 0^{- 19})

70 (70 \cdot 1 0^{6})

What is (t-1)e^{-a(t-1)}u(t-1) ?
The solution to (t-1)e^{-a(t-1)}u(t-1) is e^{-a(t-1)}u(t-1)^2