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## Algebra (all content)

### Course: Algebra (all content)>Unit 11

Lesson 20: Advanced interpretation of exponential models (Algebra 2 level)

# Interpreting change in exponential models: with manipulation

Sal analyzes the rate of change of various exponential models, where the function that models the situation needs some manipulation.

## Want to join the conversation?

• i watched this so many times and still dont get none of it • At , I repeated the last part many times yet I still don't get it, I just don't understand how 5% is the answer •  The common ratio is 1.051, which means that every time t is increased by 1, M(t) is multiplied by 1.051. Since 1.051 is larger than 0, M(t) will be increasing. 1.051 = 105.1%, and since anything multiplied by 100% is itself, M(t) will increase by 5.1% (105.1-100) every day. Hope this helps!
• At , Did Sal forget to write 1.35^5? Or was that on purpose? • can someone please explain this video in simpler terms? like a step-by-step process? thank you. • So basically what we want to find in the video is the common ratio for 1 day but we are given the common ratio for t/5 +6 days. By using the properties of exponents. we already know, Sal simplifies it in a way such that the power of the common ratio is 1. Hope you find this helpful.
• Does this mean that the initial mass of the sunfish is 1.35^5 = 4.48...? • Anyone who doesn't understand this lesson needs to go back to Unit 6 and fully master Lesson 4 of Unit 6 (may seem frustrating at first, but you will get the hang of it). • at this point the hardest part isnt the math but just reading the problem and answering in the correct units lol • Can someone give me a few examples of how to solve problems like this? Im kinda-very stuck on it...

Thankyou!🙂 • Here is an example taken from the exercise following this video:
A sample of an unknown chemical element naturally loses its mass over time.
The relationship between the elapsed time t, in days, since the mass of the sample was initially measured, and its mass, M(t), in grams, is modeled by the following function:
M(t)=900⋅(8/27)^t
The sample loses 1/3 of its mass every ____ days.
Let's look at the different parts of the equation.
900 grams is the sample's initial mass.
Multiplying that by (8/27)^t means that by the end of each day only 8/27 of the amount that was there at the beginning of that day is left. In other words, the sample loses 19/27 of its mass over the course of one day. This is more than 1/3, so it must be that the sample loses 1/3 of its mass in less than one day.
For the purpose of this exercise, we can mostly ignore the original mass, 900.
If we can change the common ratio from 8/27 to 2/3 somehow, then we will be able to find out how long it takes for the sample to lose 1/3 of its mass.
8/27=2/3^3, so:
(8/27)^t=
((2/3)^3)^t=
(2/3)^3t
This means that after about 1/3 of a day 2/3 of the mass that was there at the beginning of the day remains. In other words, the sample loses 1/3 of its mass approximately every 0.33 days.  