 # Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.) What is the relative growth rate? Use the model to estimate the world population in 1993 and to predict the population in year 2020. mSolution We measure the time t in years and let t=0 in the year 1950. We measure the population P(t) in millions of people. Then P(0)=2560 and P(0)=3040 P(t)=P(0)e^kt=2560e^kt P(10)=2560e^kt=3040 mksfmasterio 2022-09-17 Answered
Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960
Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.) What is the relative growth rate? Use the model to estimate the world population in 1993 and to predict the population in year 2020.
Solution We measure the time t in years and let t=0 in the year 1950. We measure the population P(t) in millions of people.
$P\left(0\right)=3040$
$P\left(t\right)=P\left(0\right){e}^{kt}=2560{e}^{kt}$
$P\left(10\right)=2560{e}^{kt}=3040$
This is where I have a problem.
I apply the natural logarithm to both sides of the equation.
$\mathrm{ln}2560{e}^{10k}=\mathrm{ln}3040$
I move the exponent up front. I am not sure if I am allowed to move the constant and variable.
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$\mathrm{ln}\left(2560{e}^{10k}\right)=\mathrm{ln}\left(2560\right)+\mathrm{ln}\left({e}^{10k}\right)=\mathrm{ln}\left(2560\right)+10k\mathrm{ln}\left(e\right)=\mathrm{ln}\left(2560\right)+10k$ I think you confused yourself with whether the base of the power was e or 2560e

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