Need help please

[Tara Marie]

Can someone kindly help me with the following question in bold. Am i on the right track with the formula 30=6e^kt? Please help me figure the right formula. Thanks so much.

A new disease spreads to an area. At first, 4 people contract it. Seven days later, 2 more people contract it for a total of 6. Assume the spread is exponential. What is the exponential growth rate?

1.) How many people will have contracted this disease in 30 days?

 

[Dr.Peterson]

Start at the beginning, when there are 4 people, not 6. And don't try to answer question 1 yet; before you can do that, you need to find what k is. For that, you'll be using the 6.

So, start with the equation N = 2e^(kt), which fits the initial amount. Then write an equation describing the 7-days-later case, and use that to solve for k.

Then you'll be ready to answer questions.

 

[Steven G]

To OP, Dr Peterson meant to write that N = 4e^(kt)

The reason it works is since at the start t=0 days and N, the number of people with the disease, is 4.

Note that N(0) = 4e^(k*0) = 4e^(0) = 4*1=4 which is correct, ie after 0 days there are 4 people with the disease. Now you need to find k. Use the fact that when t = 7, N=6. So solve 6 = 4e^(k*7) for k. Do not approximate k. Replace k into the formula and the find N(30).

 

[Tara Marie]

6=4e^0.879009 ??

 

[Otis]

6=4e^0.879009 ??

Hi Tara Marie. That equation is false.

Were you trying to say k = 0.879009? That value is incorrect.

Please show your work solving this equation for k:

4·e^(k·7) = 6

Thanks.

?

 

[Steven G]

e^7k = 6/4 = 3/2.

e^(kt) = (3/2)^t/7

N(t) = 4e^kt= 4(3/2)^t/7.

Now find N(30)

 

[Steven G]

6=4e^0.879009 ??

Even if what you wrote is true, you wanted to find k. So the equation you want should be in the form k = something or e^(kt) = something

 

[Tara Marie]

30=4(3/2)^t/7=
=34.78554??
I suck so bad, in any event I truly appreciate your effort to help!

 

[Otis]

… I suck so bad …

Hi. You just need more practice using logarithms and working with exponential growth/decay. (Don't feel bad, if you can't understand what Jomo did in post #6. Some of his results do not make sense.)

This is the equation we need to solve for k:

4e^(7k) = 6

Note that k is currently in the exponent position. We need to get k out of the exponent position (our goal is write k=expression, not e^k=expression). We're going to use a basic property of logarithms, to get k out of the exponent position.

We first isolate the exponential term e^(7k), on the left-hand side. We do that by dividing each side of the equation above by 4:

e^(7k) = 3/2

Now that we have the exponential term by itself, we take the natural logarithm of each side:

ln(e^[7k]) = ln(3/2)

Now we're ready to apply a basic property of logarithms. (You're expected to memorize this property and understand how to use it. One of the reasons you're struggling is because you haven't practiced enough using logarithms and properties to recall when and how to use them.)

Here is the basic property of logarithms that we need right now:

ln(b^n) = n · ln(b)

On the left-hand side, we see the natural logarithm of a power (that power is b^n). This property tells us -- when we take the logarithm of a power -- that we may move the exponent n out front as a factor, while changing the exponent on b to 1. In other words, it allows us to get n out of the exponent position.

In the equation we're trying to solve, we took the natural logarithm of a power:

ln(e^[7k])

We apply the property and write:

7k · ln(e)

The equation we're trying to solve for k now looks like this:

7k · ln(e) = ln(3/2)

You're expected to have also memorized that the expression ln(e) always represents 1 (and understand why). Therefore, we replace the expression ln(e) above with the number 1.

7k = ln(3/2)

Finish solving for k. Then replace symbol k with your result below.

N = 4e^(k·t)

You can now find N, when t is 30, by setting t=30 above and evaluating the right-hand side.

PS: Are you currently enrolled in a math class, Tara Marie?

?

 

[Steven G]

(Don't feel bad, if you can't understand what Jomo did in post #6. Some of his results do not make sense.)

EXCUSE ME! Only joking. Of course in a classroom setting I showed more of this process to my classes and no one got it. What do you think about this technique? All I did was convert the power from 7t to kt. This process gives a nice clean formula.

 

[Otis]

… What do you think about this technique? …

At this point, I think it's good for grinding gears in Tara Marie's vehicle. (You'd previously suggested that she solve for k. I'm not sure she understands the transition.)

 

[Steven G]

At this point, I think it's good for grinding gears in Tara Marie's vehicle. (You'd previously suggested that she solve for k. I'm not sure she understands the transition.)

Yeah, I did say solve for k, then I remembered it is easier and cleaner to solve for e^(kt)

 

[Otis]

… I remembered it is easier and cleaner to solve for e^(kt)

Easier for Tara Marie, to understand? Perhaps, it could be, with a modicum of segue and an iota of annotation. (I've assumed the student was shown a standard approach, in class.)

?