In a controlled breeding program at Melbourne Zoo, the
population (P) of koalas at t years is modeled by
P = P_0 * 10^(kt). Given P_0 = 20 and k = 0.3:
a. calculate the number of koalas after 2 years
b. determine when the population will be equal to 1000
thanks
To calculate the number of koalas after 2 years (a), we can use the given equation P = P_0 * 10^(kt) and substitute the values of P_0 and k.
a. Calculation:
P = P_0 * 10^(kt)
P = 20 * 10^(0.3*2)
P = 20 * 10^0.6
Calculating 10^0.6 = 3.981071705534972,
P = 20 * 3.981071705534972,
P ≈ 79.62143411169944.
Therefore, the number of koalas after 2 years is approximately 79.621.
To determine when the population will be equal to 1000 (b), we need to solve the equation P = P_0 * 10^(kt) for t.
b. Calculation:
P = P_0 * 10^(kt)
1000 = 20 * 10^(0.3t)
Divide both sides by 20 to isolate the exponential term:
50 = 10^(0.3t)
Taking the logarithm of both sides (base 10):
log10(50) = log10(10^(0.3t))
log10(50) = 0.3t * log10(10)
Simplify:
log10(50) / 0.3 = t
Using a calculator, calculate the value of log10(50) / 0.3 ≈ 4.633468455579586.
Therefore, the population will be equal to 1000 after approximately 4.633 years.
To calculate the number of koalas after 2 years, you can substitute the given values into the model and solve for P:
P = P_0 * 10^(kt)
Given P_0 = 20 and k = 0.3, the equation becomes:
P = 20 * 10^(0.3t)
To find the number of koalas after 2 years, substitute t = 2 into the equation:
P = 20 * 10^(0.3 * 2)
P = 20 * 10^(0.6)
To calculate this value, use a scientific calculator or perform the calculation step by step:
First, calculate 10^0.6:
10^0.6 ≈ 3.981
Then, multiply 20 by the result:
P ≈ 20 * 3.981
P ≈ 79.62
Therefore, after 2 years, the estimated number of koalas in the population would be approximately 79.62.
Now, let's determine when the population will be equal to 1000. To find this, we'll set up the equation:
1000 = 20 * 10^(0.3t)
We need to solve for t. Let's simplify the equation and isolate the exponential term:
10^(0.3t) = 1000 / 20
10^(0.3t) = 50
Next, take the logarithm base 10 of both sides to bring down the exponent:
0.3t = log(50) (base 10)
0.3t ≈ 1.699
Finally, divide both sides by 0.3 to solve for t:
t ≈ 5.663
So, when the population will be equal to 1000, it will occur approximately after 5.663 years.
(a) just plug in your numbers
(b) just solve 20*10^(0.3t) = 1000