Question

asdfghjkl

Answer 1

What equations have you set up to model this problem?

Answer 2

hola

Answer 3

I'd first start by deciphering everything you have

Answer 4

If you are unsure where to begin, review exponential functions ( http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/ExpDecayL.htm)


You should obtain

\[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]

Answer 5

first lets make 2010 your P-not or
\[P_0=P(0)\]

Answer 6

so at

\[P(0)=50,000\]

Answer 7

for checkerville yes

Answer 8

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

\[P=70,000e^{-.04t}\]

Answer 9

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

Answer 10

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

Answer 11

that is \[50000(1.07)^x=70000(.96)^x\]

Answer 12

essentially the same thing lol

Answer 13

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

Answer 14

@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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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