Applications of Exponential Functions

[eutas1]

Hello,

I am having troubles on how to start this very simple question... (refer to image attached).

Thank you!

 

[Dr.Peterson]

​

One thing that may be confusing you is that you can choose either a or k arbitrarily, and then determine the other. For example, we often take a to be e, which doesn't restrict possibilities at all; or you could use a^t, implicitly taking k=1. Here, since the data involve doubling, I would take a=2 to make things simple. But you could also use e, or 10, or whatever you like. So "the values of a and k" is misleading; you just need to find a pair a, k that works.

Also, be aware that P(0) could be anything; you don't need to find it.

Do you see something to do next?

 

[eutas1]

View attachment 29043​

One thing that may be confusing you is that you can choose either a or k arbitrarily, and then determine the other. For example, we often take a to be e, which doesn't restrict possibilities at all; or you could use a^t, implicitly taking k=1. Here, since the data involve doubling, I would take a=2 to make things simple. But you could also use e, or 10, or whatever you like. So "the values of a and k" is misleading; you just need to find a pair a, k that works.

Also, be aware that P(0) could be anything; you don't need to find it.

Do you see something to do next?

How can you just put random values for a?

 

[Dr.Peterson]

How can you just put random values for a?

Try it and find out! That's the best way to learn these things.

The way I usually see this taught is that a book will choose one of these options for "exponential functions":

• [imath]f(x)=Ce^{kx}[/imath] for some constant k
• [imath]f(x) = C\cdot 2^{kx}[/imath] for some constant k
• [imath]f(x) = Ca^{x}[/imath] for some constant a
• or other similar forms

I don't think I've ever seen a problem where you are asked to determine both the base and the coefficient in the exponent. Instead, they pick a fixed value for a and determine k, or the take k=1 and determine a.

The fact is that the two parameters interact. For example, [imath]f(x) = Ca^{x} = C\left(e^{\ln a}\right)^x= Ce^{(\ln a)x}[/imath], so that k is equivalent to [imath]\ln(a)[/imath].

I suggest you take k=1 and see if you can find the base. That's what is usually done when exponential functions are first introduced.

 

[eutas1]

I got a = ∛P(0) when k = 1.... which can't be right...

 

[Dr.Peterson]

I got a = ∛P(0) when k = 1.... which can't be right...

Please show how you got that, so we can help!

To see if it's right, we can plug in numbers. You have [imath]P(t) = P(0)a^t[/imath], which you're saying is [imath]P(0)\left(P(0)^{\frac{1}{3}}\right)^t[/imath]; so for t=3 you get [imath]P(3) = P(0)\left(P(0)^{\frac{1}{3}}\right)^3= P(0)\left(P(0)\right)^1 = P(0)^2[/imath]. That isn't 2P(0), which you want, so it isn't right.

 

[eutas1]

Oh wait I just made a stupid mistake...
So is this correct? (Please refer to my attached image).
I still don't get why you can just make k = 1 though... The question asks you to FIND the values of a and k, not guess one of them to get the other!

 

[Dr.Peterson]

Oh wait I just made a stupid mistake...
So is this correct? (Please refer to my attached image).

Yes, now you've got it, though you should replace k with 1 in the last line.

I still don't get why you can just make k = 1 though... The question asks you to FIND the values of a and k, not guess one of them to get the other!

But I've explained twice, this is not how this sort of question is usually asked; and you are not just guessing a value of k, you are choosing what you want to use. Normally, they choose for you! So asking for "the values of a and k" is quite misleading. I consider it to be a bad question; I'm trying to help you learn from their mistake.

I'd be very interested to see what you were taught about exponential growth laws, if this problem is not your first exposure to the concept.

Try again, but this time choose to let a = 2, which as I said is particularly suitable for this problem because it is about doubling. You'll find that the answer you get is equivalent to what you got the first way, but has a nicer form.

 

[eutas1]

Yes, now you've got it, though you should replace k with 1 in the last line.


But I've explained twice, this is not how this sort of question is usually asked; and you are not just guessing a value of k, you are choosing what you want to use. Normally, they choose for you! So asking for "the values of a and k" is quite misleading. I consider it to be a bad question; I'm trying to help you learn from their mistake.

I'd be very interested to see what you were taught about exponential growth laws, if this problem is not your first exposure to the concept.

Try again, but this time choose to let a = 2, which as I said is particularly suitable for this problem because it is about doubling. You'll find that the answer you get is equivalent to what you got the first way, but has a nicer form.

Ah, I see. Hmmmmmm okay, thank you!!!!!

 

[HocTapHay]

Your explanation is very detailed, I am looking for solutions like this, thanks a lot.

 

[Steven G]

Ca^kt
Suppose instead of the base being a you want the base to be 2.
The question is whether or not you can write a as 2 to some power. The answer is yes, since 2^log₂a=a.
We then have Ca^kt = C2^(log₂a)kt
Now let L=(log₂a)k

Now you have Ca^kt = C2^Lt

This can be done for any base of your choice.

As Dr Peterson said, you can pick a and find k.