What is the answer to A(t)=30e^{(0.02)t} ?

The answer to A(t)=30e^{(0.02)t} is t=-(\W(-3/(5A)))/(0.02)

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A(t)=30e^{(0.02)t}

A (t) = 30 e^{(0.02) t}

t = - \frac{W ( - \frac{3}{5 A} )}{0.02}

Solution steps

A (t) = 30 e^{(0.02) t}

Prepare

A (t) = 30 e^{(0.02) t}

for Lambert form:

A t e^{- 0.02 t} = 30

Rewrite the equation with

- 0.02 t = u

and

t = - 50 u

A (- 50 u) e^{u} = 30

Rewrite

A (- 50 u) e^{u} = 30

in Lambert form:

e^{u} u = - \frac{3}{5 A}

Solve

e^{u} u = - \frac{3}{5 A} : u = W (- \frac{3}{5 A})

u = W (- \frac{3}{5 A})

Substitute back

u = - 0.02 t,

solve for

t

Solve

- 0.02 t = W (- \frac{3}{5 A}) : t = - \frac{W ( - \frac{3}{5 A} )}{0.02}

t = - \frac{W ( - \frac{3}{5 A} )}{0.02}

n, x = x^{\frac{1}{n}}

8^{2 x - 1} = 15

(\frac{1}{5})^{x} = \frac{1}{25}

(7^{2})^{9} = (7^{3})^{x}

t, F = D + B a^{t}

What is the answer to A(t)=30e^{(0.02)t} ?
The answer to A(t)=30e^{(0.02)t} is t=-(\W(-3/(5A)))/(0.02)