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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 $20$th 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.

Then $P(0)=2560$ and $P(0)=3040$

$P(t)=P(0)e^{kt}=2560e^{kt}$

$P(10)=2560e^{kt}=3040$

This is where I have a problem.

I apply the natural logarithm to both sides of the equation.

$\ln 2560e^{10k}=\ln3040$

I move the exponent up front. I am not sure if I am allowed to move the constant and variable.

$10k\ \ln 2560e = \ln 3040$

I am trying to isolate $k$. How can I do that? This is more an algebra problem at this point.

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    $\begingroup$ You made a smaller error in the last step - it should be $\ln(2560)+10k = \ln(3040)$. Then subtract the $\ln$ term from the LHS to the RHS and divide by $10$. $\endgroup$
    – user208649
    Commented Oct 2, 2015 at 17:15
  • $\begingroup$ A $\LaTeX$ hint: if you precede functions like ln with a backslash, they come out in the right font. \ln becomes $\ln$ compared to ln which becomes $ln$ $\endgroup$
    – Ross Millikan
    Commented Oct 2, 2015 at 17:17
  • $\begingroup$ Can you make me aware of the rule? $\endgroup$
    – Sunny
    Commented Oct 2, 2015 at 17:17

1 Answer 1

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$\ln(2560 e^{10k})=\ln (2560) + \ln(e^{10k})=\ln (2560) +10k \ln(e)=\ln (2560) + 10k$ I think you confused yourself with whether the base of the power was $e$ or $2560e$

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  • $\begingroup$ I see now. e should always be treated as a factor. Thanks. $\endgroup$
    – Sunny
    Commented Oct 2, 2015 at 17:35
  • $\begingroup$ Your right side would be correct for $\ln((2560e)^{10k})$ (after which you could do $\ln(2560e)=\ln(2560)+1$ but the exponentiation should group more tightly than multiplication, which it does in your earlier lines. $\endgroup$
    – Ross Millikan
    Commented Oct 2, 2015 at 17:40
  • $\begingroup$ When one divides both sides by 10. Why is the right hand side read 1/10 $\endgroup$
    – Sunny
    Commented Oct 2, 2015 at 19:18
  • $\begingroup$ I don't understand the question. $1/10$ is the inverse of $10$ $\endgroup$
    – Ross Millikan
    Commented Oct 2, 2015 at 19:29
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    $\begingroup$ You are dividing both sides by $10$. It is the same as solving $2x=3$ by dividing by $2$ and getting $x=\frac 32$ $\endgroup$
    – Ross Millikan
    Commented Oct 2, 2015 at 20:34

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