Main content

### Course: Algebra 2 > Unit 8

Lesson 6: Solving exponential models# Exponential model word problem: medication dissolve

Sal solves an exponential equation in order to answer a question about an exponential model.

## Want to join the conversation?

- I understand how Sal evaluated the equation, I DON'T understand how the model originated. I'm having difficulty understanding how to use math to model ideas and concepts. Could you explain in English what the equation means? This is what I have so far, but I'm not sure if this is correct: The original dose is 20mg and this dose is reduced in the bloodstream by e (2.71828) every 48 minutes ( .80t , where t is an hour)...... is this right? This problem of using math to model ideas/concepts in the real world has been an ongoing problem for me, do you have any advice or resources you could direct me towards?(22 votes)
- You seem to be on the right track ...

However, the 0.8 means it takes**longer**than an hour (75 minutes) to get a reduction by a factor of e.

I don't have any specific resources to recommend, but I'm sure there are MOOCs (e.g. Coursera) on mathematical modeling ...

EDIT:

An example from Coursera below:

https://www.coursera.org/learn/model-thinking(19 votes)

- Can someone please explain to me why we use the natural logs here? In my mind it seems like we're altering the answer when we add them. How does the natural logs simplify things?(10 votes)
- Logs are the opposite of exponents the way multiplication is the opposite of division and addition is the opposite of subtraction. And in particular, taking log base e of e (or
`ln(e)`

) is very handy because`ln(e) = 1`

, just like`log₂2 = 1`

. After all, we can rewrite`log₂2 = 1`

in exponential form as`2^1 = 2`

.

You don't have to worry about altering the answer, as long as you follow the golden rule of equation manipulation: "Whatever you do to one side of the equation you must do to the other." Sal does, so this approach works out fine and is the easiest way to go. It's not the only way though. You could take the common log of both sides and reach the same answer, it just wouldn't be as clean a process.(10 votes)

- Why is e so useful in modelling these kind of situations? Why don't we use another number, like 2? This function could be expressed as (20)(2)^(-1.15t). I've seen answers that suggest that using e is the most 'natural' way of expressing exponential functions, but why is that?(6 votes)
- Sal made some great videos about the constant "e" that I found very intriguing. It's in the section "The Constant e and the Natural Logarithm." Here's the first video:

https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/e-and-the-natural-logarithm/v/e-through-compound-interest(6 votes)

- How do you go from Ine^(-0.8t) to -0.8t? Where did the e go? I'm confused.(4 votes)
- When you do a ln[e^(-0.8t)] it is asking for the exponent of "e" that will create e^(-08.t). The exponent is -0.8t.

Or, you can think of it as the ln cancels out the "e" leaving just its exponent.

Hope this helps.(5 votes)

- I'm having trouble comprehending this. What does t mean in -0.8t?(4 votes)
- t stands for hours.

hope this helps(4 votes)

- logbase4(4)- 3logbase2(2)

(3-logbase3(3))^3

how do you solve these 2 questions im very confused(4 votes)- Here is how:

log_4(4)-3log_2(2)

log_4(4)=1, because 4^1=4.

log_2(2)=1, because 2^1=2:

1-(3*1)

Use the order of operations: 3*1=3; 1-3=-2

log_4(4)-3log_2(2)=-2

Next, we have:

(3-log_3(3))^3

First, we should do the logarithm: log_3(3)=1:

(3-1)^3

Next, calculate what is inside the parentheses -- 3-1=2:

(2)^3

Evaluate the exponents:

(2)^3=8

(3-log_3(3))^3=8(4 votes)

- I feel like i missed something :( How sal knew the starting value?(3 votes)
- Which number are you talking about? Could you be more specific? Then, maybe I can help.(6 votes)

- Does anyone know how to do change of base?(3 votes)
- You take any (log_a) b and rewrite it as ((Log_c) b) / ((log_c) a) where c is any number that follows the rules of the bases of logs.(5 votes)

- Log base 4 2 + log base 4 (x^2 +8) =2 how do I simply this(3 votes)
- What does -.8 equate to in terms of half life? I know 20mg is the starting dosage(3 votes)
- -0.8 isn't the half life. It just implies that that after an hour, the medication gets reduced by e^(0.8)(3 votes)

## Video transcript

- [Voiceover] Carlos has
taken an initial dose of a prescription medication. The relationship between the time, between the elapsed time t, in hours, since he took the first dose
and the amount of medication, M of t, in milligrams, in
his bloodstream is modeled by the following function. Alright, in how many hours
will Carlos have 1 milligram of medication remaining
in his bloodstream? So M of what t is equal to, so
we need essentially to solve for M of t is equal to 1 milligram. Because M of t outputs, whatever
value it outputs is going to be in milligram. So let's just solve that. So M of t is, they give us a definition, it's model is an exponential
function, 20 times e to the negative 0.8 t is equal to one. So let's see, we can
divide both sides by 20 and so we will get e to the negative 0.8 t is equal to one over 20, one over 20. Which we could write as 0.05, 0.05. I have a feeling we're gonna
have to deal with decimals here regardless. And so how do we, how do we solve this? Well one way to think about
it, one way to think about it if we took, what happens
if we took the natural log of both sides? And just a remember, a
reminder, the natural log is the logarithm base e. So actually let me write
this, let me write this a little bit differently. So this is zero and that is 0.05. So I'm gonna take the natural
log of both sides, so ln, ln. So the natural log, this
says, what power do I have to raise e to, to get to
e to the negative 0.8 t? Well I've got to raise e
to the, this simplifies to negative 0.8 t. Once again natural log this thing... let me clarify ln of e
to the negative 0.8 t. This is equivalent to if
I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to, to get to e to the negative 0.8 t? Well I have to raise it to
the negative 0.8 t power. So that's why the left-hand
side simplified to this and that's going to be
equal to the natural log, actually I'll just
leave it in those terms, the natural log of 0.05, the natural log of 0.05 all of that and now we
can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8 and so t is going to be equal
to all of this business. On the left-hand side now we just have a t and on the right-hand side
we have all of this business which I think a calculator
will be valuable for. So let me get a calculator out. Clear it out and let's start with 0.05. Let's take the natural log, that's that button right over
there, the natural log. We get that value and we want
to divide it by negative .8. So divided by, divided by .8 negative. So we're gonna divide by
.8 negative is equal to, let's see they wanted us to
round to the nearest hundredth so 3.74, so it'll take 3.74, seven four hours for his dosage to go down to one milligram
where it actually started at 20 milligrams. When t equals zero it's 20 after 3.74 hours he's down in his bloodstream to one milligram. I guess his body has metabolized
the rest of it in some way.