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Mathematics 17 Online
OpenStudy (anonymous):

California Sea Otters are currently listed as 'threatened' on the Endangered Species List, which affords them special protections to help the population grow to a stable level. To be delisted, the population must reach approximately 3000 otters. In 1986 the population was at 1500 otters. In what year should the population have reached 3000 if the annual growth rate had stayed at 20%?

OpenStudy (anonymous):

OpenStudy (anonymous):

Please help me and I will give you a medal :)

OpenStudy (anonymous):

each year, you multiply your population by 1.20 (100% plus the 20%). So the population would equal: P = 1500(1.2)^t where t is in years since 1986 and P is in otters We have to solve for t OR graph it and find when it reaches 3000 OR guess and check for 3000. Which do you want to do?

OpenStudy (anonymous):

graph it I think ? becuase there i tried eliminaing t or number of year and it is almost not possible thanks for the help

OpenStudy (anonymous):

Have you studied logs?

OpenStudy (anonymous):

yes I have

OpenStudy (anonymous):

y = a^x is y = loga^ x

OpenStudy (anonymous):

but in this equation p = 3000 = y

OpenStudy (anonymous):

a= 1500(1.20) right?

OpenStudy (anonymous):

or graph it with Y1 = P and also graph Y2 = 3000 and find intersection

OpenStudy (anonymous):

okay, take the natural log or the base 10 log of both sides: ln(P) = ln(1500(1.2)^t) Then break up the multiplication with the rule of logs that does that: ln(P) = ln(1500) + ln(1.2)^t Then use the rule of logs that lets you make an exponent into a coefficient: ln(P) = ln(1500) + t ln(1.2) Then solve for t: [ln(P) - ln(1500)]/ln(1.2) = t Then evaluate at 3000: [ln(3000) - ln(1500)]/ln(1.2) = t

OpenStudy (anonymous):

p = 3000 right

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

t= 1 ?

OpenStudy (anonymous):

I get 3.8

OpenStudy (anonymous):

so the answer would be 4 years or 1990?

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