What is (e^{-kt}}{e^{-k(t+\frac{t^1)/2)}} ?

The solution to (e^{-kt}}{e^{-k(t+\frac{t^1)/2)}} is e^{(kt)/2}

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simplify (e^{-kt}}{e^{-k(t+\frac{t^1)/2)}}

simplify

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

e^{\frac{k t}{2}}

Solution steps

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

Apply exponent rule:

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

= e^{- k t - (- k (t + \frac{t ^{1}}{2}))}

Apply rule

a^{1} = a

t^{1} = t

= e^{- k t - (- k (t + \frac{t}{2}))}

Apply rule

- (- a) = a

= e^{- k t + k (t + \frac{t}{2})}

k (t + \frac{t}{2}) = \frac{3 k t}{2}

= e^{- k t + \frac{3 k t}{2}}

Join

- k t + \frac{3 t k}{2} : \frac{k t}{2}

= e^{\frac{t k}{2}}

= e^{\frac{k t}{2}}

(a x - b y)^{2} (a x + b y)^{2}

e^{- a \infty} (- \infty)

28.1 \cdot 1 0^{- 5}

(9 u) (8 u^{2})

What is (e^{-kt}}{e^{-k(t+\frac{t^1)/2)}} ?
The solution to (e^{-kt}}{e^{-k(t+\frac{t^1)/2)}} is e^{(kt)/2}