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What equations have you set up to model this problem?
hola
I'd first start by deciphering everything you have

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Other answers:

If you are unsure where to begin, review exponential functions (http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/ExpDecayL.htm) You should obtain \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]
first lets make 2010 your P-not or \[P_0=P(0)\]
so at \[P(0)=50,000\]
for checkerville yes
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 \[P=70,000e^{-.04t}\]
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
i am going to make a guess that you are supposed to use \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]
that is \[50000(1.07)^x=70000(.96)^x\]
essentially the same thing lol
divide by 50000 get \[ 91.07)^x=1.4(.96)^x\]divide by\[ (.96)^x\] get \[\left(\frac{1.07}{.96}\right)^x=1.4\]
@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
at any rate, solve via \[x=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]
say this exact same problem yesterday. they really reach for these word problems, don't they?
satellite, your equation is correct and your final solution is also correct
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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