thanks
=============================
Added by CaptainBlack
Please do not delete your questions after they have been solved.
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solutions.
thanks
=============================
Added by CaptainBlack
Please do not delete your questions after they have been solved.
It makes it more difficult for someone else to benefit from the
solutions.
You have,Originally Posted by JohnSena
$\displaystyle 25(1.06)^x=100$
Divide sides by 25,
$\displaystyle (1.06)^x=4$
Take, logarithm
$\displaystyle \log (1.06)^x=\log 4$
Thus, the exponent "comes down"
$\displaystyle x\log (1.06)=\log 4$
Thus,
$\displaystyle x=\frac{\log 4}{\log 1.06}\approx 23.8$
a)You have,Originally Posted by JohnSena
$\displaystyle 2e^{4x}=8e^{6x}$
Divide by 2,
$\displaystyle e^{4x}=4e^{6x}$
Divide by $\displaystyle e^{4x}$ thus,
$\displaystyle 1=4e^{6x-4x}=4e^{2x}$
Divide by four,
$\displaystyle 1/4=e^{2x}$
Take natural logarithm of both sides,
$\displaystyle \ln .25=2x$
Thus,
$\displaystyle x=.5\ln .25\approx -.69$
a) the rate of growth of population is proportional to the existing population.Originally Posted by JohnSena
So the population satisfies this differential equation:
$\displaystyle
\frac{dp}{dt}=\alpha p(t)
$,
which has solution:
$\displaystyle
p(t)=p(0) e^{\alpha t}
$.
Now we are told that the population grows by $\displaystyle 8\%$ per year, so
after one year the population is $\displaystyle 1.08$ times its original value so:
$\displaystyle
p(1)=1.08 p(0)=e^{\alpha}p(0)
$,
so: $\displaystyle \alpha=\log(1.08)$ (thats a natural log by the way).
hence:
$\displaystyle
p(t)=p(0) e^{\log(1.08) t}=1000 e^{\log{1.08}t}
$.
b) The rate of population growth is a constant so the population satisfiesOriginally Posted by JohnSena
the differential equation:
$\displaystyle
\frac{dp}{dt}=70
$,
which has solution:
$\displaystyle
p(t)=70t+p(0)
$,
plugging in the given initial population gives:
$\displaystyle
p(t)=70t+1000
$.
RonL
Missing assumption needed to solve this is that the rate of growth of theOriginally Posted by JohnSena
population is proportional to the existing population. Hence the population
satisfies the diffrenetial equation:
$\displaystyle
\frac{dp}{dt}=\alpha p
$,
which has solution:
$\displaystyle
p(t)=p(0) e^{\alpha t}
$
Now we are told that the population doubles in 8 hours so:
$\displaystyle
p(8)=2p(0)=p(0) e^{8\alpha}
$,
so: $\displaystyle 2=e^{8\alpha}$, or: $\displaystyle \alpha=\frac{\log(2)}{8}
$.
Therefore:
$\displaystyle
p(t)=p(0)e^{\frac{\log(2)}{8}t}
$.
Now at time $\displaystyle t_5$ the population is 5 times its original value so:
$\displaystyle
p(t_5)=5p(0)=p(0)e^{\frac{\log(2)}{8}t_5}
$,
so:
$\displaystyle
5=e^{\frac{\log(2)}{8}t_5}
$
or laking logs:
$\displaystyle
\log(5)=\frac{\log(2)}{8}t_5
$,
or:
$\displaystyle
t_5=\frac{8\log(5)}{\log(2)}
$
RonL