thanks

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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.

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- May 21st 2006, 07:27 PMJohnSenaNeed solutions for some easy calc problems
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. - May 21st 2006, 07:31 PMThePerfectHackerQuote:

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,

$\displaystyle x=\frac{\log 4}{\log 1.06}\approx 23.8$ - May 21st 2006, 07:34 PMThePerfectHackerQuote:

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,

$\displaystyle x=.5\ln .25\approx -.69$ - May 21st 2006, 07:57 PMJohnSena
damn that was quick thanks

still confused on how to figure out part b and the other problems though, if anybody could help with those that'd be awesome - May 21st 2006, 11:00 PMCaptainBlackQuote:

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}

$. - May 21st 2006, 11:03 PMCaptainBlackQuote:

Originally 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 - May 21st 2006, 11:17 PMCaptainBlackQuote:

Originally 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 - May 26th 2006, 08:15 AMCaptainBlackQuote:

Originally Posted by**JohnSena**

It makes it more difficult for someone else to benefit from the

solutions.

RonL