SOLUTION: A certain type of skin wound heals according to the function given by N(t)= N0e-0.125t , where N is the number of cm2 of unhealed skin t days after the injury, and N0 is the number

Question 233600: A certain type of skin wound heals according to the function given by N(t)= N0e-0.125t , where N is the number of cm2 of unhealed skin t days after the injury, and N0 is the number of cm2 of the original wound. How long, to the nearest day, will it take for 75% of the wound to heal?

Please see below for further break downs
N0= The 0 is slightly lower than the N
-0.125t= It is squared to the N0e
cm2= the 2 is squared here
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
is what I believe your formula looks like after putting it through algebra.com formula generator.

Just put three { in front and three } in back of the formula to see what algebra.com formula generator will make your formula look like.

Example:

formula is: N[t]= N[0]*e^(-0.125*t)

Algebra.com formula generator makes it look like

Anyway, your answer looks like it is 11 days to the nearest integer.

The full answer shows up as 11.09035489

Here's how it works:

N[t] is what you are looking for.

If you want to know how long it will take for 75% of the wound to heal, then you need to translate that into the percent of the wound that didn't heal which would make that 25%, because the formula is set up to show you the amount of skin remaining to be healed.

If you allow the original wound to be 1 square cm, then 25% of that would be .25 square cm.

N[0] would be 1
N[t] would be .25

Your formula of:

N[t]= N[0]*e^(-0.125*t) would become:

.25 = 1*e^(-0.125*t)

This becomes:

.25 = e^(-0.125*t)

Take the natural log of both sides to get:

ln(.25) = ln(e^(-0.125*t))

This becomes:

ln(.25) = -.125*t*ln(e)

since ln(e) = 1, this becomes:

ln(.25) = -.125*t

divide both sides by -.125 to get:

t = ln(.25)/(-.125)

solve for t to get:

t = 11.09035489

I confirmed the answer is good by plugging into the original equation and solving.

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