Respuesta :
Answer:
It is going to take 17.65 years for the population to double.
Step-by-step explanation:
The population of the deer is after t years is given by the following equation
[tex]P(t) = P_{0}(1 + r)^{t}[/tex]
In which [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.
A deer population grows at a rate of four percent per year. This means that [tex]r = 0.04[/tex]
How many years will it take for the population to double?
This is t when [tex]P(t) = 2P_{0}[/tex]
[tex]P(t) = P_{0}(1.04)^{t}[/tex]
[tex]2P_{0} = P_{0}(1.04)^{t}[/tex]
[tex](1.04)^{t} = 2[/tex]
Here, we apply the log 10 to both sides of the equation.
It is important to note the following property of logarithms.
[tex]\log{a^{t}} = t\log{a}[/tex]
[tex]\log{(1.04)^{t}} = \log{2}[/tex]
[tex]t\log{1.04} = 0.3[/tex]
[tex]0.017t = 0.3[/tex]
[tex]t = \frac{0.3}{0.017}[/tex]
[tex]t = 17.65[/tex]
It is going to take 17.65 years for the population to double.
Answer:it will take approximately 18 years
Step-by-step explanation:
A deer population grows at a rate of four percent per year. The growth rate is exponential. We would apply the formula for exponential growth which is expressed as
A = P(1 + r/n)^ nt
Where
A represents the population after t years.
n represents the period of growth
t represents the number of years.
P represents the initial population.
r represents rate of growth.
From the information given,
A = 2P
P = P
r = 4% = 4/100 = 0.04
n = 1
Therefore
2P = P(1 + r/n)^ nt
2P/P = (1 + 0.04/1)^1 × t
2 = (1.04)^t
Taking log of both sides to base 10
Log 2 = log1.04^t = tlog1.04
0.3010 = t × 0.017
t = 0.3010/0.017 = 17.7 years
