Question

7 years ago the population was at 1720 and today only 950 of the birds are alive. Once the population drops below 150, the situation will be irreversible. When will this happen?

Answer 1

Answer:

This will happen in 28.79 years.

Explanation:

The population, after t years, is given by the following equation:

$P(t) = P_{0}(1 + r)^{t}$

In which $P_{0}$ is the initial population and r is the rate of change. If r is positive, the population increases. Otherwise, if r is negative, the population decreases.

7 years ago the population was at 1720 and today only 950 of the birds are alive.

This means that $P_{0} = 1720$ and $P(7) = 950$. With this, we can find r. So

$P(t) = P_{0}(1 + r)^{t}$

$950 = 1720(1 + r)^{7}$

$(1+r)^{7} = \frac{950}{1720}$

$(1+r)^{7} = 0.5523$

To isolate 7, i apply the seventh root to both sides. so

$\sqrt[7]{(1+r)^{7}} = \sqrt[7]{0.5523}$

$1 + r = 0.9187$

$r = 0.9187 - 1$

$r = -0.0813$

So

$P(t) = 1720(1 - 0.0813)^{t}$

$P(t) = 1720(0.9187)^{t}$

Once the population drops below 150, the situation will be irreversible. When will this happen?

This is t when P(t) = 150. So

$P(t) = 1720(0.9187)^{t}$

$150 = 1720(0.9187)^{t}$

$(0.9187)^{t} = \frac{150}{1720}$

$(0.9187)^{t} = 0.0872$

It is important to know that:

$\log{a^{t}} = t\log{a}$

So we apply log to both sides

$\log{(0.9187)^{t}} = \log{0.0872}$

$t\log{0.9187} = \log{0.0872}$

$t = \frac{\log{0.0872}}{\log{0.9187}}$

$t = \frac{-1.0595}{-0.0368}$

$t = 28.79$

This will happen in 28.79 years.