Hello, bergausstein!
2. The population of a community is known to increase
at a rate proportional to the population at any time t.
We have: .\frac{dP}{dt} \:=\:kP \quad\Rightarrow\quad \frac{dP}{P} \:=\:k\,dt
Integrate: .\ln|P| \:=\:kt+C
. . P \:=\:e^{kt+c} \:=\:e^{kt}\cdot e^c \:=\:e^{kt}\cdot C
Hence: .P(t) \:=\: Ce^{kt}
When t = 0,\,P = P_o, initial population.
. . P_o \:=\:Ce^0 \quad\Rightarrow\quad C \,=\,P_o
Therefore: .P(t) \;=\;P_oe^{kt}
If the population is doubled in 5 years,
how long will it take to triple?
When t = 5,\;P=2\!\cdot\!P_o
We have: .2\!\cdot\!P_o \:=\:P_oe^{5k} \quad\Rightarrow\quad e^{5k} \:=\:2
. . 5k \:=\:\ln2 \quad\Rightarrow\quad k \:=\:\tfrac{1}{5}\ln2
Hence: .P(t) \:=\:P_oe^{(\frac{1}{5}\ln2)t} \:=\:P_o\left(e^{\ln2}\right)^{\frac{1}{5}t}
Then: .P(t) \;=\;P_o\!\cdot\!2^{\frac{1}{5}t}When will P(t) = 3\!\cdot\!P_o\,?
.3\!\cdot\!P_o \:=\:P_o\!\cdot\!2^{\frac{1}{5}t} \quad\Rightarrow\quad 2^{\frac{1}{5}t}\:=\:3
. . \ln\left(2^{\frac{1}{5}t}\right) \:=\:\ln(3) \quad\Rightarrow\quad \tfrac{1}{5}t\ln(2) \:=\:\ln(3)
. . t \:=\:\frac{5\ln(3)}{\ln(2)} \:=\:7.924...
About 7.9 years.