Exponential Growth and Decay

Sol 1 Since $C=5$, we have that $y=Ce^{kt}=5e^{kt}$.

Since $y=7$ when $t=11$, $5e^{11k}=7$; so $e^{11k}=7/5$. Taking natural logarithms of both sides gives $11k=\ln 7/5$, so $k=\frac{1}{11}\ln 7/5$.

Substituting back in the formula for $y$ gives

$y=5e^{(\frac{1}{11}\ln 7/5)t}=5(e^{\ln 7/5})^{t/11}=5(7/5)^{t/11}$, so

$y=10\Rightarrow 5(7/5)^{t/11}=10\Rightarrow (7/5)^{t/11}=2$.

Taking natural logarithms on both sides gives

$\frac{t}{11}\ln 7/5=\ln 2$, so $t\ln 7/5=11\ln 2$ and

$$t=\frac{11\ln 2}{\ln 7/5} \mbox{ hours.}$$

Sol 2 Since $C=5$, we have that $y=Ce^{kt}=5e^{kt}$.

Since $y=4$ when $t=24$, $5e^{24k}=4$; so $e^{24k}=4/5$. Taking natural logarithms of both sides gives $24k=\ln 4/5$; so $k=\frac{1}{24}\ln 4/5$.

Substituting back in the formula for $y$ gives

$y=5e^{(\frac{1}{24}\ln 4/5)t}=5(e^{\ln 4/5})^{t/24}=5(4/5)^{t/24}$.

To find the half-life, we can set $y=(1/2)(5)$ and solve for $t$:

$y=5/2\Rightarrow 5(4/5)^{t/24}=5/2\Rightarrow (4/5)^{t/24}=1/2$.

Taking natural logarithms on both sides gives

$\frac{t}{24}\ln 4/5=\ln 1/2$, so $t\ln 4/5=24\ln 1/2$ and

$$t=\frac{24\ln 1/2}{\ln 4/5}=\frac{24(-\ln 2)}{-\ln 5/4}=\frac{24\ln 2}{\ln 5/4} \mbox{ years.}$$

Sol 3 We know that $C=15,000$, so $y=Ce^{kt}=15,000e^{kt}$.

Since $y=25,000$ when $t=10$, we have that

$15,000e^{10k}=25,000$ and therefore $e^{10k}=25/15=5/3$.

Taking natural logarithms on both sides yields

$10k=\ln 5/3$, so $k=\frac{1}{10}\ln 5/3$ and therefore

$y=15,000e^{(\frac{1}{10}\ln 5/3)t}=15,000(e^{\ln 5/3})^{t/10}=15,000(5/3)^{t/10}$.

When the population has increased by 40%, it will be equal to $15,000+(.40)(15,000)=(1.40)(15,000)$; so

$y=1.4(15,000)\Rightarrow 15,000(5/3)^{t/10}=1.4(15,000)\Rightarrow (5/3)^{t/10}=1.4$.

Taking natural logarithms gives

$\frac{t}{10}\ln 5/3=\ln 1.4$, so $t\ln 5/3=10\ln 1.4$ and hence

$$t=\frac{10\ln 1.4}{\ln 5/3}\mbox{ years.}$$

Sol 4 Since $k=.05$, $y=Ce^{kt}=Ce^{.05t}$. To find the time required for the amount to triple, we can set $y=3C$ and then solve for $t$:

$y=3C\Rightarrow Ce^{.05t}=3C\Rightarrow e^{.05t}=3 \Rightarrow .05t=\ln 3$, so

$t=(\ln 3)/(.05)=20\ln 3$ years.

Sol 5 Here $C=500$, so $y=Ce^{kt}=500e^{kt}$.

Since $y=450$ when $t=3$, we have that $500e^{3k}=450$, so

$e^{3k}=450/500=9/10 \Rightarrow 3k=\ln 9/10\Rightarrow k=\frac{1}{3}\ln 9/10$.

Thus $y=500e^{(\frac{1}{3}\ln 9/10)t}=500(e^{\ln 9/10})^{t/3}=500(9/10)^{t/3}$.

The number of words remembered has decreased by 40% when $y=500-(.40)(500)=(.60)(500)$, so

$y=(.60)(500)\Rightarrow 500(9/10)^{t/3}=(.60)500\Rightarrow (9/10)^{t/3}=.6$;

and taking natural logarithms on both sides gives

$\frac{t}{3}\ln 9/10=\ln(.6)$. Therefore we get that

$$t=\frac{3\ln (.6)}{\ln (.9)}\mbox{ weeks.}$$

Sol 6 Since the number of people infected doubles every 5 weeks, and since $8=2^3$, the number of people infected will increase by a factor of 8 [that is, double 3 times] in $3(5)=15$ weeks.

Sol 7 Let $y(t)$ be the number of grasshoppers in the town (in thousands) after t days.

Then we know that

$$\frac{y(15)}{y(0)}=\frac{y(30)}{y(15)}$$

since the proportional amount of change in any 15-day time period is the same, so

$(y(15))^2=(y(0))(y(30))=8(18)=144$ and therefore

$y(15)=12$ thousand grasshoppers.

Sol 8A Since $y=2880$ when $t=3$ and $y=5120$ when $t=5$, we have that

$Ce^{3k}=2880$ and $Ce^{5k}=5120$.

Dividing the second equation by the first gives

$$\frac{Ce^{5k}}{Ce^{3k}}=5120/2880,$$

so $e^{2k}=512/288=128/72=16/9$. Taking the square root of both sides gives

$e^k=4/3$, so substituting back in the first equation gives

$C=\frac{2880}{e^{3k}}=\frac{2880}{(e^k)^3}=\frac{2880}{(4/3)^3}=2880(3/4)^3$, so

$C=2880(27/64)=(45)(27)=1215$ bugs.

Sol 8B Making a time-shift, let $t=0$ correspond to the time when there were 2880 bugs.

Then $y=Ce^{kt}$ where $C=2880$, so $y=2880e^{kt}$; and $y=5120$ when $t=2$.

Therefore $2880e^{2k}=5120$, so $e^{2k}=512/288=128/72=16/9$.

Taking square roots of both sides gives $e^k=4/3$, so substituting back in the formula for $y$ gives

$y=2880e^{kt}=2880(e^k)^t=2880(4/3)^t$.

Since we shifted the time by 3 weeks, the initial number of bugs is given by

$y(-3)=2880(4/3)^{-3}=2880(3/4)^3=2880(27/64)=45(27)=1215$.

Sol 9 For state A, we have that $y=9e^{kt}$ (in millions) where

$y=9+(.03)(9)=(1.03)(9)$ when $t=1$; so letting $t=1$ in the formula for $y$ gives

$9e^k=(1.03)(9)\Rightarrow e^k=1.03\Rightarrow y=9e^{kt}=9(e^k)^t=9(1.03)^t$.

For state B, we have that $y=11e^{kt}$ (in millions) where

$y=11+(.02)(11)=(1.02)(11)$ when $t=1$; so letting $t=1$ in the formula for $y$ gives

$11e^k=(1.02)(11)\Rightarrow e^k=1.02\Rightarrow y=11e^{kt}=11(e^k)^t=11(1.02)^t$.

Setting the expressions for the populations of the two states equal to each other and solving for $t$, we get

$9(1.03)^t=11(1.02)^t \Rightarrow \frac{(1.03)^t}{(1.02)^t}=\frac{11}{9}\Rightarrow$

$(1.03/1.02)^t=11/9\Rightarrow (103/102)^t=11/9\Rightarrow t\ln (103/102)=\ln (11/9)$ and

$$t=\frac{\ln (11/9)}{\ln (103/102)}\mbox{ years.}$$

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