A =Ao ^(-0.35n): the area of a healing wound

[hollerback1]

This is a very complex problem I don't get:

The normal healing of a wound can be modeled by A = A₀^-0.35n, where A is the area of the wound in square centimeters after n days. After how many days is the area of the wound half of its original size, A₀?

Please explain to me what this problem is about

 

[galactus]

I believe you forgot the e in your formula.

\(\displaystyle \L\\A=A_{0}e^{-0.35n}\)

Find out after how many days(n) the wound is half its original size.

It's original size is \(\displaystyle A_{0}\). Therefore, half of its original size is

\(\displaystyle \frac{A_{0}}{2}\).

So, we have \(\displaystyle \L\\\frac{A_{0}}{2}=A_{0}e^{-0.35n}\)

Divide both sides by \(\displaystyle A_{0}\):

\(\displaystyle \L\\\frac{1}{2}=e^{-0.35n}\)

Can you solve for n now?.

 

[hollerback1]

1/2=e^-0.35n after that, what do i do now?

 

[stapel]

after that, what do i do now?

Then you solve by whatever method(s) you've learned in class for solving exponential equations. Probably, you'll take the natural log of both sides.

Eliz.

 

[hollerback1]

I just tried natural log form, and it said nonreal answers on the calculator.

I'm confused, please help

 

[stapel]

I just tried natural log form, and it said nonreal answers on the calculator.

You have a calculator that does symbolic (rather than numeric) operations...?

Please reply showing what you tried. Thank you.

Eliz.

 

[hollerback1]

this is what i did....

1/2=e^-0.35n I did ln(-0.35n)/ln(1/2)

 

[stapel]

this is what i did....

1/2=e^-0.35n I did ln(-0.35n)/ln(1/2)

So you haven't yet learned in class how to solve exponential equations...?

The general technique is to take the log of both sides. Any base will do, but since the base you already have is the natural exponential, the natural log would be a good choice. No calculator is involved in this process.

Here is an example:

. . . . .Solve 3 = 5^4x

. . . . .1) Take any log of each side.

. . . . .ln(3) = ln(5^4x)

. . . . .2) Apply the log rules you've memorized.

. . . . .ln(3) = 4x ln(5)

. . . . .3) Solve for the variable.

. . . . .ln(3) / [4 ln(5)] = x

Follow that process with your exercise. If you get stuck, please reply showing your steps.

(If you need further instruction, please specify such, and we'll be glad to find some online lessons, so you can learn about this topic.)

Thank you.

Eliz.