Question

1. The population of toads on an pond can be modeled by the equation where t  Pt=120e^-.698t is the number of years since 2000.

1. What was the population in the year 2000?
2. Describe what is happening to the population. Is it growing or shrinking?

c.When will there be only 20 toads on the pond?

Answer 1

Solution:

According to the question:

P(t) = 120e -0.6986

(a)

$P(t)=P_0~e^{-rt}\\\\Where\\P_0~~initial~population~\\r~~rate\\t~~time$

Population in 2000 is 120

(b)

The population is shrinking because model equation e power has a negative constant factor.

At times increases the population is decreasing.

(c)

P(t)=20 t=?

$\\20=120e^{-0.698t}\\\mathrm{Divide\:both\:sides\:by\:}120\\e^{-0.698t}=\frac{1}{6}\\\\\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)\\\\\ln \left(e^{-0.698t}\right)=\ln \left(\frac{1}{6}\right)\\\\\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\\\\-0.698t\ln \left(e\right)=\ln \left(\frac{1}{6}\right)\\\\\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1\\\\-0.698t=\ln \left(\frac{1}{6}\right)\\\\\mathrm{Solve\:}\:-0.698t=\ln \left(\frac{1}{6}\right):\quad t=\frac{500\ln \left(6\right)}{349}\\\\t=2.5669$

After 2.567 years the population will be only 20 toads on the pound.