Sal solves an exponential equation in order to answer a question about an exponential model.
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- 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?(17 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 ...
An example from Coursera below:
- 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?(5 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 = 1in 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.(6 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?(3 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:
- I feel like i missed something :( How sal knew the starting value?(2 votes)
- Does anyone know how to do change of base?(1 vote)
- 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.(3 votes)
- logbase4(4)- 3logbase2(2)
how do you solve these 2 questions im very confused(1 vote)
- Here is how:
log_4(4)=1, because 4^1=4.
log_2(2)=1, because 2^1=2:
Use the order of operations: 3*1=3; 1-3=-2
Next, we have:
First, we should do the logarithm: log_3(3)=1:
Next, calculate what is inside the parentheses -- 3-1=2:
Evaluate the exponents:
- Just wanted to say that I accidently clicked no on the " Was the video helpful?" but it really was. The only problem I have is, what if we have to write the answer as a natural logarithm?(1 vote)
- If you have to write the answer as a natural logarithm, then just don't do the last step where you evaluate the logarithm. Just leave it as t=(ln(1/20))/-0.8. I think that should be acceptable.(1 vote)
- [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.