Log application word problem.

[ballet432]

The rate of population entering = G(x) = g'(x) = 4(.96)^X thousands/year
The rate of population leaving = D(x) = f'(x) = (ln(x+1)) / (ln(3.95)) thousands/year

a) find the total number of ppl who entered the state over 0 < = X < = 12 (<= stands for less than or equal to)
b) find expression using definite integral which shows the # of ppl in the state at any time t, where 0 < = X < = 20 if the population of the state at time t = 0 is 108,000 ppl
C) when is the rate of the population entering equal to the rate leaving (approx what year)
D)what is the population of the city when x =12 (ie. t=12) and when x=18 (ie. t=18)


I am really confused on this problem, any help would be appreciated.

 

[soroban]

Hello, ballet432!

I have a start on this problem . . .

$\displaystyle \text{The rate of population entering: }\;E'(x) \:=\: 4(0.96)^x\,\text{ thousands/year}$

$\displaystyle \text{The rate of population leaving: }\;L'(x) \:=\: \frac{\ln(x+1)}{ \ln(3.95)}\,\text{ thousands/year}$

$\displaystyle \text{(a) Find the total number who entered the state over }0 \leq x \leq 12$

$\displaystyle E(x)\bigg]^{12}_0 \;=\;\int^{12}_0 E'(x)\,dx \;=\;4\int^{12}_0 (0.96)^x\,dx \;=\;4\ln(0.96)\!\cdot\!(0.96)^x\,\bigg]^{12}_0$

. . . . . . $\displaystyle =\;4\ln(0.96)\bigg[(0.96)^{12} - (0.96)^0\bigg] \;=\;0.063239841$

$\displaystyle \text{About 63 people entered the state.}$

$\displaystyle \text{(b) Find an expression, using definite integrals, which shows the population of the state}$
$\displaystyle \text{at any time }t\text{, where }0 \leq x \leq 20\,\text{ if the population of the state at time }x = 0\text{ is }108,000.$

$\displaystyle P(x) \;=\;\int^{20}_04(0.96)^x\,dx \;- \int^{20}_0\frac{\ln(x+1)}{\ln(3.95)} \,dx + 108,000$

C) When is the rate of the population entering equal to the rate leaving (approx what year)

$\displaystyle \text{You must solve: }\;4(0.96)^x \;=\;\frac{\ln(x+1)}{\ln(3.95)}\,\text{ by some numerical method.}$

 

[BigGlenntheHeavy]

$\displaystyle a) \ G(12) \ = \ 4000\int_{0}^{12}(.96)^{t}dt \ \dot= \ 37,949 \ people \ entered \ the \ state \ in \ a \ 12 \ year \ period.$

$\displaystyle b) \ P(x) \ = \ 4000\int_{0}^{x}(.96)^{t}dt-\frac{1000}{ln(3.95)}\int_{0}^{x}ln|t+1|dt \ + \ 108,000, \ 0 \ \le \ x \ \le \ 20.$

$\displaystyle c) \ 4(.96)^{t} \ = \ \frac{ln|t+1|}{ln(3.95)}, \ t \ \dot= \ 16.15 \ \dot= \ 16th \ year, \ about \ 2,069 \ people \ came \ and \ went.$

$\displaystyle d) \ P(12) \ = \ 4000\int_{0}^{12}(.96)^{t}dt \ - \ \frac{1000}{ln(3.95)} \int_{0}^{12}ln|t+1|dt \ + \ 108,000 \ \dot= \ 130,412 \ people.$

$\displaystyle d) \ P(18) \ = \ 4000\int_{0}^{18}(.96)^{t}dt \ - \ \frac{1000}{ln(3.95)} \int_{0}^{18}ln|t+1|dt \ + \ 108,000 \ \dot= \ 131,370 \ people.$

$\displaystyle Shades \ of \ Detroit?$