City Population Growth: A Unique Algebra Problem

[Tompson Lee]

Hey, I found an interesting algebra problem combined with log. It was quite an unique one so I wanted to share it with you guys.

Problem: There is a city where the population increases at a constant rate. At the end of 2016, the population of the city was 2 times larger than the population 15 years ago (the end of 2001). Find as a percentage, how much did the population at the end of 2007 increase compared to the end of 2001? Use 0.12 as log1.32 and 0.30 as log2.

I believe that this is quite difficult one to solve. It took me over an hour to solve it. Please tell me your guys way to solve it since that was my purpose to post this not to know the answer.

Thanks

Solution:
[YOUTUBE]sTK8Tu32VMA[/YOUTUBE]

 

[HOI]

Hey, I found an interesting algebra problem combined with log. It was quite an unique one so I wanted to share it with you guys.

Problem: There is a city where the population increases at a constant rate.

So, letting "x(t)" be the population at time t (in years), dx/dt= kx where x is the constant rate of increase. Then dx/x= kdt so ln(x)= kt+ C so x(t)= C'e^{kt} where C'= e^C. Taking t= 0 to be 2001, the population in 2001 to be x0, C'= x0. The population t years after 2001 is x0e^{kt}.

At the end of 2016, the population of the city was 2 times larger than the population 15 years ago (the end of 2001).

Note that "2 times larger than" is NOT "twice as large". "2 times larger than" is "three times as large".

At the end of 2016, 15 years after 2001, x(15)= x0e^{15k}= 3x0 so e^{15k}= 3. We can use the natural logarithm to say that k= ln(3)/15. x0e^{kt}= x0e^{ln(3)t/15}

Find as a percentage, how much did the population at the end of 2007 increase compared to the end of 2001? Use 0.12 as log1.32 and 0.30 as log2.

2007 is 6 years after 2001 so the population is x0e^{6ln(3)/15}= x0 e^{2ln(3)/5}. The ratio with the population at 2001 is x(5)/x_0= e^{2ln(3)/5}. This can be written as (e^{ln(3)})^{2/5)= 3^{2/5}.

I have no idea why this would want you to use the common logarithm to solve. Does anyone still use that? But, anyway, with x= 3^{2/5}, log(x)= (2/5)log(3)= (2/5)(0.48) (to two decimal places)= 0.19.

I believe that this is quite difficult one to solve. It took me over an hour to solve it. Please tell me your guys way to solve it since that was my purpose to post this not to know the answer.

Thanks

Solution:

Since you are given "log(2)" rather than "log(3)", perhaps this solution is interpreting "two times larger" as "twice as large". If so, that is wrong!