The population N(t) (in millions) of a country t years after 1980 may be approximated by the formula N(t) = 213e^0.0101t. When will the population be twice what it was in 1980?
Population t years after 1980 is given by:
𝒹ₚ/𝒹ₜ = N(t) = 213 e⁰·⁰¹⁰¹ ᵗ
integrating it we get,
P= 213·C·e⁰·⁰¹⁰¹ ᵗ ...(i)
[Where C is a constant of integration]
when t=0 i.e. in 1980,
P= 213C
when population of 1980 is doubled,
i.e P= 2P= 2×213·C
Now from equation (i)
2×213·C = 213·C·e⁰·⁰¹⁰¹ ᵗ
2=e⁰·⁰¹⁰¹ ᵗ
taking log on both sides,
log 2 = 0.0101 t
t = (log 2)/0.0101= 29.54 years
what a lot of work, when you just want to find when
e^0.0101t = 2
Actually this was all to explain the person who posted the problem.
I could have skipped some lines.
I don't know what his/her level of education is, so I just used my own method to solve it and also explained it well enough to be understood.
Thank you for your suggestion.
To find out when the population will be twice what it was in 1980, we need to set up an equation and solve for the desired value of t.
The population in 1980 is given by N(0) since t represents the number of years after 1980. Plugging t = 0 into the formula, we get:
N(0) = 213e^(0.0101*0) = 213e^0 = 213 * 1 = 213 million.
To find the year when the population will be twice this value, we set up the equation:
2 * N(0) = N(t).
Substituting the value of N(0) into the equation, we get:
2 * 213 = 213e^(0.0101t).
Now we can solve for t. Divide both sides of the equation by 213:
2 = e^(0.0101t).
To isolate e^(0.0101t), take the natural logarithm (ln) of both sides:
ln(2) = ln(e^(0.0101t)).
Using the logarithmic property ln(e^x) = x, we can simplify further:
ln(2) = 0.0101t.
Now divide both sides by 0.0101 to solve for t:
t = ln(2) / 0.0101.
Using a calculator, we can evaluate this expression to find the value of t:
t ≈ 68.32.
Therefore, the population will be twice what it was in 1980 approximately 68.32 years after 1980, which corresponds to the year 2048.