fruitbat Group Title asdfghjkl one year ago one year ago

1. binarymimic

What equations have you set up to model this problem?

2. Outkast3r09

hola

3. Outkast3r09

I'd first start by deciphering everything you have

4. binarymimic

If you are unsure where to begin, review exponential functions (http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/ExpDecayL.htm) You should obtain $50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}$

5. Outkast3r09

first lets make 2010 your P-not or $P_0=P(0)$

6. Outkast3r09

so at $P(0)=50,000$

7. Outkast3r09

for checkerville yes

8. Outkast3r09

lets rephrase what your equation is $P=ce^{kt}$ where p is your population , c is a constant unknown atm and k is your rate. t is time. if we put in P(0) you get $50,000=ce^{0}=c$ $P=50,000e^{kt}$ since we know the rate at which it increases 7% $P=50,000e^{.07t}$ next doing for checker ville you get $P=70,000e^{-.04t}$

9. Outkast3r09

At P1=P2 is when one will exceed the other and the other will be below soooo $70,000e^{-.04t}=50,000e^{.07t}$ solve for t

10. satellite73

i am going to make a guess that you are supposed to use $50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}$

11. satellite73

that is $50000(1.07)^x=70000(.96)^x$

12. Outkast3r09

essentially the same thing lol

13. satellite73

divide by 50000 get $91.07)^x=1.4(.96)^x$divide by$(.96)^x$ get $\left(\frac{1.07}{.96}\right)^x=1.4$

14. satellite73

@Outkast3r09 yeah it is the same except that i am thinking that because it say 7% per year it is using $$(1.07)^x$$ rather than $$e^{.07x}$$ but i could be wrong

15. satellite73

at any rate, solve via $x=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}$

16. satellite73

say this exact same problem yesterday. they really reach for these word problems, don't they?

17. binarymimic

satellite, your equation is correct and your final solution is also correct

18. Outkast3r09

nah it's because i meant to put the rate as 1.07 and .96 so you'd get the same thing