For the below problem, I solved it as you suggested and got
k=.0486
When I did 300,000 = 122,000e^.0486t
I got about 18.5 years for when the sewer system would need to be updated.
Next, for how many people will the new system have to accommodate for the system to last 15 years before needing to be updated again,
Should I use 300,000 for the p(o)?
pre-calc - bobpursley, Friday, January 8, 2010 at 9:08pm
Population is usually exponential growth
P(t)=P(o)e^kt
P(o)=122000
find k from
180000=122000 e^kt
log of both sides
ln(180000)=ln(122000)+ kt
then t=8, solve for k
see if you can do it from there. find t when P(t)=300,000
then find p(15+the t you just found).
The city planners of bakersfield, CA are studying the current and future needs of the city. The sewage system they have in place now can support 300,000 residents. The current population of the city is 180,000, up from 122,000 eight years ago. Assuming that the population will continue to row at this rate, when will the sewer system need to be updated? How many people will the new have to accomodate for the system to last 15 years before needing to be updated again?
Ok, so I know that 180,000 - 122,000 is 58,000 so 58,000/122,000 = 47.5%
increas in eight years. I'm not sure where to go from here.
Should I be using the formula for exponential growth?
yes, use 300,000 as P(o) and t as 15
Yes, you should be using the formula for exponential growth, which is P(t) = P(o)e^(kt).
To find the value of k, you can use the equation P(t) = P(o)e^(kt) and plug in the values you have. In this case, P(t) is 180,000 and P(o) is 122,000. After substituting these values, you can solve for k.
ln(180,000) = ln(122,000) + kt
Solve for k:
k = (ln(180,000) - ln(122,000))/t
Since t is given as 8 years, substitute this value into the equation to find k. You should get k ≈ 0.0486.
Now that you have obtained the value of k, you can use it to solve the first question: when will the sewer system need to be updated?
The equation P(t) = P(o)e^(kt) can again be used, this time with t as the variable to solve for. Given P(o) as 122,000 and P(t) as 300,000, you can plug in these values along with the value of k you just calculated.
300,000 = 122,000e^(0.0486t)
Solve for t by taking the natural logarithm of both sides:
ln(300,000) = ln(122,000) + 0.0486t
Solve for t using the obtained value for k. It appears that you got t ≈ 18.5 years, which is correct.
Finally, to answer the second question of how many people the new system will have to accommodate for the system to last 15 years before needing to be updated again, you should use P(o) = 300,000. This is because you want to find the population at the end of 15 years after the initial population of 300,000.
So, substitute P(o) = 300,000 into the exponential growth equation and plug in the value of t you just found (18.5 years + 15 years). By calculating P(t), you will obtain the number of people the new system will have to accommodate.
P(t) = P(o)e^(kt)
P(t) = 300,000e^(0.0486(18.5 + 15))
Calculate P(t) to get the desired population.