Question

A quantity with an initial value of 8500 decays exponentially at a rate such that the

Answer 1

Answer:

$4250 = 8500 e^{50 k}$

And we can divide both sides by 8500 and we got:

$\frac{1}{2} = e^{50 k}$

And we can solve for k using natural log

$k = \frac{ln(\frac{1}{2})}{50}= -0.0138629$

And we have the model give by:

$P(t) = 8500 e^{-0.01386294361 t}$

And if we replace t =97 we got:

$P(t=97) = 8500 e^{-0.01386294361 *97}= 2215.240$

Step-by-step explanation:

For this case we can use the proportional model given by:

$\frac{dP}{dt}= kP$

And we can rewrite the expression like this:

$\frac{dP}{P} = k dt$

If we integrate both sides we got:

$ln P = kt +C$

If we apply exponentials in both sides we got:

$P = e^{kt +C}$

And we can rewrite the expression like this:

$P(t) = P_o e^{kt}$

Where P represent the population and t the time in years since the starting value

For this case we have that $P_o = 8500$

And after t = 5*10 = 50 years the value is the half or 8500/2 = 4250

So we can use this condition and we have:

$4250 = 8500 e^{50 k}$

And we can divide both sides by 8500 and we got:

$\frac{1}{2} = e^{50 k}$

And we can solve for k using natural log

$k = \frac{ln(\frac{1}{2})}{50}= -0.0138629$

And we have the model given by:

$P(t) = 8500 e^{-0.01386294361 t}$

And if we replace t =97 we got:

$P(t=97) = 8500 e^{-0.01386294361 *97}= 2215.240$