Algebra to determine values

[MathStudent21]

Hey,

I've recently received a Grade 12 Maths B assignment and I've been stuck on this question for a few days now... I would really appreciate some help on this one if possible, THANKS IN ADVANCE!

It isknown that the function of the population of the Stapylococcus Aureus (Psa) is usually ab exponential of the form Psa = ce^dt (where c and d are the constants).
The following data was found. (Imagine its in a table !)

Time (t) (hours) 2 9
Population (Psa) (millions) 100 8

The question asks: By using an algebraic approach, determine the values of c and d (to 3 decimal places). Hence show the population is 206e^-0.361t.

206e-0.361t
PSA = ce^dt
100=ce^d2
C=100/e^d2

8=ce^d9
C=8/e^d2

100/e^d2 = 8/e^d9
Ln100/e^d2 = Ln8/e^d9
Ln100+d9 = Ln8+d9
D = Ln8-Ln100/7
D = -0.36082

Now to find C we I subbed that into the equations C=100/e^d2 & C=8/e^d9
I got C=100/e^-0.36082x2 = 205.78
and C=8/e^d9 = 246.37
This is where i get stuck^
I have no idea how to get one value for c
Any help would be much appreciated

 

[tkhunny]

100=ce^d2
C=100/e^d2

Please don't switch variable names mid-stream. "C" is not "c".

100/e^d2 = 8/e^d9
Ln100/e^d2 = Ln8/e^d9

Why did you do that? There is much simpler simplification before you'll need logarithms.

100/8 = (e^2d)/(e^9d)

50/4 = e^(2d - 9d)

25/2 = e^(-7d)

Now use the logarithms

Ln100+d9 = Ln8+d9

This is just careless. You tried to do three things at once and managed to lose your "d2"

Give it another go.