What is the answer to N(t)=300(5/4)^t ?

The answer to N(t)=300(5/4)^t is t=-(\W(-(300ln(5/4))/N))/(ln(5/4))

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N(t)=300(5/4)^t

N (t) = 300 (\frac{5}{4})^{t}

t = - \frac{W ( - \frac{300 l n ( \frac{5}{4} )}{N} )}{ln ( \frac{5}{4} )}

Solution steps

N (t) = 300 (\frac{5}{4})^{t}

Prepare

N (t) = 300 (\frac{5}{4})^{t}

for Lambert form:

Nt e^{- l n (\frac{5}{4}) t} = 300

Rewrite the equation with

- ln (\frac{5}{4}) t = u

and

t = - \frac{u}{ln ( \frac{5}{4} )}

N (- \frac{u}{ln ( \frac{5}{4} )}) e^{u} = 300

Rewrite

N (- \frac{u}{ln ( \frac{5}{4} )}) e^{u} = 300

in Lambert form:

e^{u} u = - \frac{300 ln ( \frac{5}{4} )}{N}

Solve

e^{u} u = - \frac{300 ln ( \frac{5}{4} )}{N} : u = W (- \frac{300 ln ( \frac{5}{4} )}{N})

u = W (- \frac{300 ln ( \frac{5}{4} )}{N})

Substitute back

u = - ln (\frac{5}{4}) t,

solve for

t

Solve

- ln (\frac{5}{4}) t = W (- \frac{300 ln ( \frac{5}{4} )}{N}) : t = - \frac{W ( - \frac{300 l n ( \frac{5}{4} )}{N} )}{ln ( \frac{5}{4} )}

t = - \frac{W ( - \frac{300 l n ( \frac{5}{4} )}{N} )}{ln ( \frac{5}{4} )}

322 = 2^{a}

2 e^{l n (\frac{x}{3})} = 2

0.25 = 0. 5^{x}

2 (3^{4 x - 5}) + 4 = 11

\frac{50}{1 + e ^{- x}} = 4

What is the answer to N(t)=300(5/4)^t ?
The answer to N(t)=300(5/4)^t is t=-(\W(-(300ln(5/4))/N))/(ln(5/4))