A certain country's population P(t), in millions, t years after 1980 can be approximated by P(t) = 2.495(1.019)^t. Find the doubling time.
solve
4.990 = 2.495(1.09)^t
2 = 1.09^t
log 2 = tlog1.09
t = log2/log1.09 = 8.043
48,6 yr
58.4 yr
36.8 yr
or
73.7 yr
I should really have my eyes checked soon, lol
I saw 1.09 instead of 1.019
so last line
t = log2 / log 1.019 = 36.8 years
To find the doubling time, we need to determine the time it takes for the population to double from its initial value.
In this case, the initial population is P(0) = 2.495 million, as t represents the years after 1980.
Let's set up an equation to find the doubling time, where P(t) will be equal to double the initial population:
2P(0) = P(t)
2 * 2.495 = 2.495(1.019)^t
Now we can solve this equation to find the value of t.
Divide both sides of the equation by 2.495:
2 * 2.495 / 2.495 = (1.019)^t
2 = (1.019)^t
To get the value of t, we need to take the logarithm of both sides of the equation. Since the base of the exponential term is 1.019, using the logarithm base 1.019 will simplify the equation.
log₁.₀₁₉(2) = t
To find the value of log₁.₀₁₉(2), we can use the logarithmic property that states:
logᵦ(x) = log(y) / log(β)
In this case, β is 1.019, x is 2, and y is the base, 10:
log₁₀(2) / log₁₀(1.019) = t
Using a calculator, we can evaluate the logarithms:
log₁₀(2) ≈ 0.30103
log₁₀(1.019) ≈ 0.00798
Thus,
t ≈ 0.30103 / 0.00798
t ≈ 37.73
Therefore, the doubling time for the population is approximately 37.73 years.