## anonymous 3 years ago asdfghjkl

1. anonymous

What equations have you set up to model this problem?

2. anonymous

hola

3. anonymous

I'd first start by deciphering everything you have

4. anonymous

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. anonymous

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

6. anonymous

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

7. anonymous

for checkerville yes

8. anonymous

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. anonymous

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. anonymous

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

11. anonymous

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

12. anonymous

essentially the same thing lol

13. anonymous

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. anonymous

@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. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

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

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