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### Course: Get ready for Algebra 2>Unit 4

Lesson 6: 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?

• 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?
• I know this is kind of late but I have a method that worked for me, however it involved a calculator. I used the geometric and arithmetic explicit equations and just plugged in the values from the problem into the graphing calculator. Then I looked into the table on the graphing calculator and found my answer.
• Is it possible to do this with an equation instead of graphing it each time? Thanks :)
• It is, using somthing called log
• Is there a faster way to solve a problem like this?
• Samuel you could could graph it and record the intercept of the two lines but this way is simpler because in order to graph the equation you would have to make a table anyway.

Hope I helped,
Xavier Jalem
• So I was thinking instead of going through each x-y pairs to check when the excess happens, how about writing the inequality:

Let m = # of months

10,000 + 5000*m < 500 * 2^m

how can I solve for m here? Do I not have enough math tools to solve this inequality? If not, can you show me how to solve, perhaps using system of equations instead?
• 5000(2+m)<500*2^m
10(2+m)<2^m

The last equation is the furthest I have been able to simplify with both your equation and someone else's. I'm not sure if with all the math tools learned from this point in math and prior it'd be possible to further simplify this inequality, but I am curious about your thought on using system of equations. What 2 (or more) equations are you suggesting we use? (I have seen this idea come up a lot so I'm quite curious)
• Is there another way to do this? Some equations like this don't have the exponential function pass the linear function for a long time. I would like to know if there is a faster way to do this.
• you can try to create like A=10,000+5,000(n) or something like that. I don't think there's a specific equation but you can create your own based on the information given. Also if there's anything wrong with the equation please don't flame me T_T. I made it in like 15 seconds and just typed it in.
(1 vote)
• Sal made a mistake. he accidentally multiplied by 4 instead of 2 XD