# Population increases with annual rate problem

1. Feb 23, 2009

Population of Mexico. In 2006 the population of
Mexico was 107.4 million. If Mexico’s population continues
to grow at an annual rate of 1.43%, then the population
in 2028 will be 107.4(1.0143)14 <-------- is an exponent

a) Find the predicted population in 2020 to the nearest
tenth of a million people.

Not sure how to start

2. Feb 23, 2009

### HallsofIvy

Staff Emeritus
Re: Stuck

That value for 2028 doesn't make sense to me. There are 2028- 2006= 22 years between 2006 and 2028. If the population increases by 1.43% per year, then each year it is 1.0143 larger than the previous year. That should be 1.014322.

Last edited: Feb 24, 2009
3. Feb 24, 2009

Re: Stuck

This is what I told the instructor that there is a type and this does not make sense, he insisted that I look for help with my classmates.
Oh well I share your dilema.
Thank you for your response I will see what the rest of the class came up with.
Thank you once more I thought I was going retarded LOL

4. Feb 24, 2009

### fball558

Re: Stuck

i know its a little late, but i got an answer that makes some sence. if anything at least this might clear it up for a test or something.
i use y=Ce^(kt) C= inital population k= rate t= time y= final population.
work through first time to find k
so y=Ce^(kt) would give

this is final pop. inital pop 2028-2006 = 22years
131018987.268=107400000e^(22k) k = .00903555

then go through again this time solve for y and using k from above and new time (14 years)
y=107400000e^(.00903555*14) y = 121882548.523 cant have .523 people so round down

5. Feb 25, 2009

Re: Stuck

Thank you guys for all the help , I will be a regular from now on on this site < I love it

I will let you know the answer as soon as I get my paper back