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### Course: Algebra 2 > Unit 7

Lesson 3: Advanced interpretation of exponential models- Interpreting change in exponential models: with manipulation
- Interpret change in exponential models: with manipulation
- Interpreting change in exponential models: changing units
- Interpret change in exponential models: changing units
- Exponential models: FAQ

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# Interpreting change in exponential models: changing units

Sal analyzes the rate of change of various exponential models for different time units by manipulating the functions that model the situations.

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- I noticed that about half way through the video, when the table is filled out, the initial value of 315 was written as 3.5. I suspect this was a mistake and the value Sal meant to say was 315 the whole time?(47 votes)
- If you are watching in fullscreen, you can exit fullscreen and see that there is a correction box at the bottom right-hand corner that lets you know he messed up. Happens all the time. EDIT: Just realized this is 7 years late...(16 votes)

- Given the phrasing of the question, wouldn't the answer be 1.01? I'm looking to see if this question comes up in the quiz, and if we are marked wrong because of putting in too many decimal places. Sal didn't prep very well for this one.(26 votes)
- Yes. If you round the answer as requested by the instructions, 1.0058 would become 1.01.(26 votes)

- What I did was said that if in 10 years it went up 6%, then yearly that means it goes up 0.6%, which is close to the final answer of 0.58%. Was what I did acceptable?(11 votes)
- No, because it's not a linear function. The error in reasoning can be seen more clearly if we use a more exaggerated example. You can't say that if in one year you get 100 percent increase (or double the original), then in 10 years you'll get 1000 percent, or 10 times as much. Try doubling a number 10 times and see the difference.(15 votes)

- The first value of A(t) in the table is 315, as it goes on the value becomes 3.5. Is this a mistake?(12 votes)
- Yes, Sal slipped up. There’s usually a little box in the bottom right corner with corrections like that, though you can’t see the box if you’re in fullscreen.(1 vote)

- Using this logic, can we solve compound interest problems involving fractional periods?(5 votes)
- Yes. You can manipulate the exponent t to however your situation is. For example, if it happens every year, you can put (t/365) for the exponent. Likewise, if it's every month, you can put (t/12) — however,
**be wary of how the question uses the variable**.

Hopefully that helps !(5 votes)

- So, when I tried to model the function in terms of years instead of decades, intuitively I thought that it should look something like this:

a_years(t) = 315 * (1.06)^((1/10)t)

because in order for a_years to output an equivalent answer to a_decades, you should need to input 10 times more input into a_years. 1 year is a tenth of a decade, after all.

but Sal's model is slightly different

a_years(t) = 315 * (1.0058)^t

The first model is correct, right?(2 votes)- Both models are correct, because

1.06^(0.1𝑡) = (1.06^0.1)^𝑡 ≈ 1.0058^𝑡(7 votes)

- Where did he get the 3.5?(4 votes)
- That was a mistake. He meant to write 315.(5 votes)

- in3:57in the video, he didn't round to two decimal places. why?(5 votes)
- he later mentions it but doesn't correct it(2 votes)

- Khan's answer is wrong. Answer is 1.01 because that's 1.0058 rounded to 2dp.(5 votes)
- He stated that the whole answer is arguably correct. I see your reasoning in the blatant wording of the question, but for the sake of learning the topic itself I think it would be best to focus on the answer instead of the rounding.

So you may be right with this detail, but do you understand the whole point of the video?(1 vote)

- A(t)=315(1.06)^t

A(t)=315(1.06)^(0.1t/0.1)

A(t)=315(1.06)^(0.1t x (1/0.1))

A(t)=315(1.06^(1/0.1))^0.1t

A(t)=315(1.06^10)^0.1t

why doesn't this work, because i thought this is showing the increase for every 0.1t i.e every year?(2 votes)- Starting with the original expression: A(t) = 315 * (1.06)^t

A(t) = 315 * (1.06)^(0.1t/0.1) - This step is correct, you are simply dividing the exponent 0.1t by 0.1, which is the same as multiplying it by 1.

A(t) = 315 * (1.06)^(0.1t x (1/0.1)) - This step is still correct, as multiplying by (1/0.1) is equivalent to dividing by 0.1.

A(t) = 315 * (1.06^(1/0.1))^0.1t - Here's where the issue occurs. You've applied the exponent rule (a^(bc) = (a^b)^c) to move the 0.1t from the base to the exponent. However, this rule only applies when the exponent is a constant. In your case, 0.1t is not a constant; it depends on the value of t. So, you cannot directly move it to the exponent.

A(t) = 315 * (1.06^10)^0.1t - Following from the incorrect step above, you end up with this expression, which isn't the correct representation.

To show the increase every 0.1t (every year), you need to express the growth factor (1.06) correctly in terms of the yearly growth. The correct expression would be:

A(t) = 315 * (1.06^(0.1t))

In this expression, (0.1t) represents the number of years, and raising 1.06 to the power of (0.1t) gives you the factor by which the population increases every year. This way, you are correctly accounting for the yearly growth rate and its cumulative effect over time.

Now, the expression A(t) = 315 * (1.06^(0.1t)) represents the increase in the population every 0.1t (every year) based on a yearly growth rate of 6%.(3 votes)

## Video transcript

- [Voiceover] The amount
of carbon dioxide, CO2, in the atmosphere increases rapidly as we continue to rely on fossil fuels. The relationship between
the elapsed time, t, in decades, let me highlight that, because that's not a typical unit, but, in decades, since CO2
levels were first measured, and the total amount of
CO2 in the atmosphere, so the amount of CO2,
A, of sub decade of (t), in parts per million, is modeled by the following function. So, the amount of CO2 as a function of how many decades have passed is going to be this, so t is in decades in this model right over here. Complete the following sentence about the yearly rate of change, the yearly rate of change in the amount of CO2 in the atmosphere. Round your answer to two decimal places. Every year, the amount of CO2 in the atmosphere
increases by a factor of? If they said every decade, well, this would be pretty straightforward. Every decade, you increase t
by one, and so you're going to multiply by 1.06 again. So every decade, you
increase by a factor of 1.06. But what about every year? Now, I always find it
helpful to make a bit of a table, just so we can
really digest things properly. So I'll say t, and I'll say A of t. So, when t is equal to zero, so at the beginning of our study, well, 1.06 to the zeroth
power is just going to be one, you have 315 parts per million. So, what's a year later? So, a year later is going
to be a tenth of a decade. Remember, t is in decades. So a year later is 0.1 of a decade. So, 0.1 of a decade later, what is going to be the
amount of carbon we have? Well, it's going to be 315 times 1.06 to the 0.1 power. And what is that going to be? Let's see, if we... so 1.06 to the, so to the 0.1 power, I didn't have to actually
use the parentheses there, is equal to 1.0058, I'll just take it with that. 1.0058. So this is the same thing as 3.5 times 1.0058. And I should say approximately equal to, I did a little bit of rounding there. So, after another year,
so now now we're at t equals 0.2, we're at
two-tenths of a decade. Where're we going to be? We're going to be at 3.5 times 1.06 to the 0.2, which is the same thing as 3.5 times 1.06 to the 0.1, and then that raised to the second power. So we're going to multiply by this 1.06 to the one-tenth power again, or we're going to multiply
by 1.0058 a second time. Another way to think about it, if we wanted to reformulate this model in terms of years, so per year of t, it's going to be 315
and now our common ratio wouldn't be 1.06, it would
be 1.06 to the .1 power, or 1.0058, and then we would raise that. Now t would be in years now. Here, it is in decades. And I can say approximately, since this is rounded a little bit. And so, every year, the amount of CO2 in the atmosphere
increases by a factor of, I could say 1.06 to the 0.1
power, but if I'm rounding my answer to two decimal places, the, well, we're going to increase by 1.0058, in fact, they should, the
increase is by a factor of, they should, I'm guessing they want more than two decimal places. Well, anyway, this is arguably,
this right over here is five significant digits. But anyway, I'll leave it there.