# Exponential Growth: Modeling growth in months vs years

• B
Gold Member
Say I have a function that represents the population growth of a certain country that can be written as ##f\left(x\right)=1.25\left(1.012\right)^t##, where t is in years. I can graph this function and it will look a certain way exponentially.

I've looked at a ton of examples, and they're all modeled in years. One thing that I tried is to graph the function ##f\left(x\right)=1.25\left(1.012\right)^{t/12}## to model the graph in months rather than years. I did get a different graph which doesn't increase nearly as fast as the original. Why would the t in years appear to grow faster than t in months? In this population model, wouldn't the growth be consistent throughout the whole year?

What I don't understand here (assuming this is how you would graph the function in terms of months), is when we have an exponential function to the power of t, is it always in years? If we want to model a time frame in anything other than years, do we need to manipulate t to reflect this? Or would we just say t is in months, not years, from the start?

scottdave
Homework Helper
I think you mean to say "f(t) =" rather than f(x), since there is no "x" in your right hand side.

This represents a slow growth rate (1.2% per year). The way it is set up, it is like compounding annually. Putting t/12 in the exponent you will get to see how it
progresses after a number of months. Putting in ## 1.012^{t/12} ## with t = 12 will yield ## 1.012^{12/12} = 1.012##, just as ## 1.012^{t} ## would give the same answer with t = 1 (year)

• opus
Gold Member
I think you mean to say "f(t) =" rather than f(x), since there is no "x" in your right hand side.

This represents a slow growth rate (1.2% per year). The way it is set up, it is like compounding annually. Putting t/12 in the exponent you will get to see how it
progresses after a number of months. Putting in ## 1.012^{t/12} ## with t = 12 will yield ## 1.012^{12/12} = 1.012##, just as ## 1.012^{t} ## would give the same answer with t = 1 (year)
Yes I meant to say ##f\left(t\right)##. Thank you.
So when I look at the graph of the function with the exponent as ##\frac {t}{12}##, the entire graph that is shows is the growth model of a single month? And if I have the exponent simply as t, it will show the graph over an entire year?

And to follow your example with ##1.012^\frac {t}{12}##, and plugging in the value of 12 into t, that would be 12 months out of 12 months or one year, which is the same as just t. If I were to plug in, say, 9 into t rather than 12, that would show we the growth of the population over the entire course of 9 months, or over the single ninth month (in this case September)?

mfb
Mentor
The function calculates the value after time t. If you plug in t=1 in the first one you get the value after 1 year. If you plug in t=1 in the second formula you get the value after 1 month - that will show much less growth than a full year, of course. If you plug in t=9 in the second formula you get the value after 9 months (i.e. after September).

To compare the two formulas, you should do this at equal times, e.g. after 10 years, or t=10 for the first one and t=120 (months) for the second. Clearly 120/12=10, so you get the same result in both cases.