What is the answer to solvefor t,a(t)=a_{0}*e^{-0.025t} ?

The answer to solvefor t,a(t)=a_{0}*e^{-0.025t} is t=(\W(\frac{a_{0})/(40a))}{0.025}

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solvefor t,a(t)=a_{0}*e^{-0.025t}

solve for

t, a (t) = a_{0} \cdot e^{- 0.025 t}

t = \frac{W ( \frac{a _{0}}{40 a} )}{0.025}

Solution steps

a (t) = a_{0} e^{- 0.025 t}

Prepare

a (t) = a_{0} \cdot e^{- 0.025 t}

for Lambert form:

a t e^{0.025 t} = a_{0}

Rewrite the equation with

0.025 t = u

and

t = 40 u

a \cdot 40 u e^{u} = a_{0}

Rewrite

a 40 u e^{u} = a_{0}

in Lambert form:

e^{u} u = \frac{a _{0}}{40 a}

Solve

e^{u} u = \frac{a _{0}}{40 a} : u = W (\frac{a _{0}}{40 a})

u = W (\frac{a _{0}}{40 a})

Substitute back

u = 0.025 t,

solve for

t

Solve

0.025 t = W (\frac{a _{0}}{40 a}) : t = \frac{W ( \frac{a _{0}}{40 a} )}{0.025}

t = \frac{W ( \frac{a _{0}}{40 a} )}{0.025}

3 (4^{x}) + 4^{x - 1} = 52

4^{2 x} + 12 = 2^{2 x + 3}

4^{3 x} = 3 2^{x - 1}

2 5^{x - 1} + 100 = 5^{2 x - 1}

2^{2 x + 1} = 2^{x + 2}

What is the answer to solvefor t,a(t)=a_{0}*e^{-0.025t} ?
The answer to solvefor t,a(t)=a_{0}*e^{-0.025t} is t=(\W(\frac{a_{0})/(40a))}{0.025}