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## Integrated math 1

### Course: Integrated math 1>Unit 14

Lesson 4: Exponential vs. linear growth over time

# Exponential vs. linear growth over time

If we compare linear and exponential growth, we will see that over time, *any* exponential growth will surpass *any* linear growth, no matter how steep it is.

## Want to join the conversation?

• Are there an algebraic equation that can represent the growth over time? Can we use that equation to figure out the answer? • This method may save you some time:
Construct the equations of the two functions (if you don't know how, go to exponential expressions on the same topic ).
Using these two equations, find their values after one year, two years, three years, etc. When the one function exceeds the other, you have found it!
If it's a long problem, then first try some key points, like 3, 5, 7 and 9 -this would make your time more targeted and you will know what to look for.
Hope it helped a bit.
• I don't understand why Sal didn't express the rates of growth in terms of functions. I tried taking that approach to solve a practice exercise on Kahn Academy, but my results don't match up with the ones on the website.

Here is the practice exercise:

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Sheila is a wildlife biologist. Her daily task is to count the number of wild turkeys and white-tail deer in a large game reserve.

- Sheila counts 12 wild turkeys by the first hour after sunrise, and the cumulative number of turkeys she has counted increases by approximately 40 percent each hour.

Sheila counts 18 white-tail deer by the first hour after sunrise, and she counts 10 deer each hour after that.

In which hour after sunrise will Sheila's cumulative count of the turkeys first exceed the cumulative count of the deer?

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I will represent the exponential growth in turkeys with f(x) = 12 times 1.4^x

I will represent the linear growth in deer with f(x) = (18)x + 10

Using those functions, I calculate that by 2 hours after sunrise, there will be 23.52 turkeys (since 12 times 1.4^2 = 23.52) and 46 deer (since 18 times 2 + 10 = 46). However, when I ask Khan Academy for hints on this practice problem, I am told that there are 17 turkeys and 28 deer by 2 hours after sunrise.

I don't understand why my functions did not produce the correct results. I also don't understand why the use of functions is not encouraged to get results. • David - Per your request, here's my take.
Your functions are assuming that the clock starts after she does here initial count. You can't make that assumption. You need to account for the 1st hour as noted in the other responses the you received.

For Turkeys
There is 12 turkeys that increase by 40% after the end of the 1st your.
Let x = times. Your exponent needs to be x-1 so that the 1st hour is not increased by 40%. You want only hours 2 and above to increase by 40%. Thus, the function should be:
T(x) = 12*1.4^(x-1)
T(1) = 12*1.4^(1-1) = 12*1.4^0 = 12*1 = 12
T(2) = 12*1.4(2-1) = 12*1.4 = 16.8, or when rounded 17 turkeys.

For Deer
There is 18 deer that increase by 10 each hour after the end of the 1st your. The 18 is your starting point. The 10 is your slope. But, like with the other function, you need to adjust time to account for the 1st hour.
Let x = times. You multiply 10 by x-1 so that the 1st hour does not increase by 10. You want only hours 2 and above to increase by 10. Thus, the function should be:
D(x) = 18+10(x-1)
D(1) = 18+10(1-1) = 18+10(0) = 18+0 = 18
D(2) = 18+10(2-1) = 18+10(1) = 18+10 = 28

Hope this helps.
• how do i solve this?

A(t) = 10,000 + 5,000t
B(t) = 500 * 2^t

500 * 2^t > 10,000 + 5,000t / :500
= (500 * 2^t) / 500 > (10,000 + 5,000t ) / 500
= 2^t > 20 + 10t
from here i am stuck and i am not sure how to solve it. • This lesson doesn't expect you to know how to solve that algebraically. I don't know how to do it either. A(t) is a linear equation. B(t) is an exponential equation. The answer you want is the value of t that makes the two equations equal (which corresponds to the intersection of the two graphs). How do you solve for t when one of them is the exponent and one is not? I don't know. If you Google "finding intersection of linear and exponential functions," you will find numerical methods (not algebraic) for solving this kind of problem to an arbitrary precision. I'm just reviewing Algebra so I'm not watching many videos, but if he didn't, Sal should have mentioned this problem (because any thoughtful student of math would wonder the same thing as you).
• Is there an equation that can be used to solve this quicker, or do you have to draw the table? • Is there a faster way to solve a problem like this? • Is it cheating If I use these videos to get answers for the practices?
(1 vote) • Is there any simplest way except using table and graph? like using some formulas • There's got to be a more efficient way to do this besides making a table.. right?
(1 vote) • TB,

Graphing both offerings might be more efficient for some students. To some extent, the table is used to demonstrate the reality of what happens month-to-month and to give a simple explanation of what each offer means.

The equation for Company A's offer is

f(t) = 10,000 + 500t

The equation for Company B's offer is

f(t) = 500*2^t

The question is "For which monthly payment will Company B's payment first exceed Company A's payment?"

We are looking for the lowest whole-number value of t that will make the following true

500*2^t > 10,000 + 500t

At this point in the course, it is likely that more students have the skills to work with a table or create a graph than to solve that equation.  