Solving a growth rate/time problem: 332 people on October 2016 increasing by 19.5%...

[Whitehartlane]

Hi guys, I am just asking if there's any way I can solve the following problem with an equation (I did it the long way but it's too time consuming).

Data 1: 332 people on October 2016 increasing by 19.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

Data 2: 300 people on October 2016 increasing by 25.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

In what month and year does Data 2 eclipse Data 1?

Long answer I got July 2017. I tried to make it into an inequality but I got April 2018 which isn't right.

 

[Ishuda]

Hi guys, I am just asking if there's any way I can solve the following problem with an equation (I did it the long way but it's too time consuming).

Data 1: 332 people on October 2016 increasing by 19.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

Data 2: 300 people on October 2016 increasing by 25.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)

In what month and year does Data 2 eclipse Data 1?

Long answer I got July 2017. I tried to make it into an inequality but I got April 2018 which isn't right.

What have you been taught about exponential growth problems? If you can't remember, you should be able to work it out: Take the following problem starting at time zero with population of p₀ population increasing by rate r percent every t₀ periods. For each period we need to multiply the initial population 1+r/100. In an amount of time t, there are t/t₀ periods of length t₀. So at the end of time t, the population P is p₀ multiplied by 1+r/100 t/t0 times or
P = p₀ (1+r/100)^t/t₀

For your problem, the first data set has p₀ = 332 r = 19.5% and t₀=3 if we measure time in months starting from the beginning of Oct.
P₁(t) = 332 (1.195)^t/3
So what is P₂(t) and at what time t does P₂(t) become larger than P₁(t).

 

[Whitehartlane]

332*1.195^n = 300*1.255^n

Solve for n

That'll give you the number of "3month periods".
Convert that to months.

I found it very difficult to solve for n so I used Wolfram Alpha which gave me this figure: n = 0.243694

It doesn't seem right to me.

 

[JeffM]

Hi guys, I am just asking if there's any way I can solve the following problem with an equation (I did it the long way but it's too time consuming).Data 1: 332 people on October 2016 increasing by 19.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)Data 2: 300 people on October 2016 increasing by 25.5% every 3 months (January 2017 --> April 2017 ---> July 2017 ---> October 2017 etc.)In what month and year does Data 2 eclipse Data 1?Long answer I got July 2017. I tried to make it into an inequality but I got April 2018 which isn't right.

\(\displaystyle 300 * 1.255^n > 322 * 1.195^n \implies\)\(\displaystyle log(300) + n * log(1.255) > log(322) + n * log(1.195).\)

The least integer value of n possible is the number of months from the start date. So if you get an answer like 9.1, the least integer is 10.