Logistic growth model for US population predictions

[ineedhelpnow]

b)
$2.75(11e^{-10k}+1)=30$
$11e^{-10k}+1=10.91$
$11e^{-10k}=9.91$
$e^(-10k}=0.90083$
$-10k=ln(0.90083)$
$k=0.010444$

c)
$P(110)=\frac{30}{11e^{(-0.010444)*110}+1}$
put that into wolframalpha and you get 6.69 billion

$P(210)=\frac{30}{11e^{(-0.010444)*210}+1}$
put that into wolframalpha and you get 13.5 billion

d)
$3.5=\frac{30}{11e^{-0.01044t}+1}$
solve that for t and you get 24 so 2014

did i do them right?

 

[MarkFL]

b) Let's use:

$$11e^{-10k}+1=\frac{120}{11}$$

$$11e^{-10k}=\frac{109}{11}$$

$$e^{-10k}=\frac{109}{121}$$

$$e^{-k}=\left(\frac{109}{121}\right)^{\frac{1}{10}}$$

Now substituting into the population function, we obtain:

$$P(t)=\frac{30}{11\left(\frac{109}{121}\right)^{\frac{t}{10}}+1}$$

c) Now, in the year 2100, when $t=110$, we find:

$$P(110)=\frac{30}{11\left(\frac{109}{121}\right)^{11}+1}\approx6.67$$

Now, recall that we chose as units, 100 million people, and so this equates to about 667 million people. We should suspect something is wrong if our answer exceeds the carrying capacity of 3 billion.

And, in the year 2200, when $t=210$, we find:

$$P(210)=\frac{30}{11\left(\frac{109}{121}\right)^{21}+1}\approx13.47$$

This equates to about 1.347 billion people.

d) Let $P(t)=3.5$ and solve for $t$:

$$3.5=\frac{30}{11\left(\frac{109}{121}\right)^{\frac{t}{10}}+1}$$

$$3.5\left(11\left(\frac{109}{121}\right)^{\frac{t}{10}}+1\right)=30$$

$$11\left(\frac{109}{121}\right)^{\frac{t}{10}}=\frac{53}{7}$$

$$\left(\frac{109}{121}\right)^{\frac{t}{10}}=\frac{53}{77}$$

$$t=10\frac{\ln\left(\frac{53}{77}\right)}{\ln\left(\frac{109}{121}\right)}\approx36$$

Thus, the year is 2026

 

[ineedhelpnow]

so solving for k doesn't work?

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and where did 120/11 come from?

 

[MarkFL]

so solving for k doesn't work?

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and where did 120/11 come from?

Solving for $k$ is fine, but recall the problem to which you linked...I had kind of forgotten that we can just solve for $e^{-k}$ instead of having to go through using the rules of exponents/logs just to get back to where we would be if we just use $e^{-k}$.

The 120/11 is the exact value of the approximation 10.91 that you used, by dividing 30 by 2.75:

$$\frac{30}{2.75}=\frac{3000}{275}=\frac{25\cdot120}{25\cdot11}=\frac{120}{11}$$