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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3

Lesson 10: Topic D: Lessons 23-24: 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?
• 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
• 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?
• 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.
• 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?
• I feel like i missed something :( How sal knew the starting value?
• Which number are you talking about? Could you be more specific? Then, maybe I can help.
• Does anyone know how to do change of base?
• 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.
• How do you go from Ine^(-0.8t) to -0.8t? Where did the e go? I'm confused.
• 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.
• Log base 4 2 + log base 4 (x^2 +8) =2 how do I simply this
• I'm having trouble comprehending this. What does t mean in -0.8t?
• t stands for hours.
hope this helps
• logbase4(4)- 3logbase2(2)
(3-logbase3(3))^3
how do you solve these 2 questions im very confused
• 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