SOLUTION: Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1. I have done the following but I

Question 718152: Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1.
I have done the following but I am not convienced it is the right path...
y=ae-bx therefor e-bx=y/a ... take log e of both sides ... loge e-bx = loge y/a
but logee=1 therefor -bx=loge y/a ... move the x over to get -b=x loge y/a
then enter values for x and y from example 1... -b= (0.6) loge(1000/a)
Is this the correct path to take for solving this?? Any advice would be greafull appreciated. Thank you.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
when x=0.6, y=1000 and when x=1.8, y=1

No matter what you do, you are going to have to solve a system of two equations with two unknowns, a and b. Although what you've done is not an invalid path, I'm not convinced that it makes the problem any easier.

We'll start by correcting an error in your work and then finishing the problem. Later we'll look at ways you might find as good or better than what you've been trying.

Your work is find up to:

But to "move x over" you have to divide by x, not multiply it:

Substituting in the values for x and y we get two equations:

and

Using the Substitution Method on this system, we would substitute one equation's expression for -b in for the other equation's -b:

Then we would solve this for a. Then we would use this value for a and one of the earlier equations (which has a and b) to solve for b.

Although the path you were trying to take eventually would have worked (if there was no error), here's a couple of alternatives you might find easier and/or faster. They both start with substituting in the x's and y's into the original equation:

giving us:

and

which simplify to:

and

From here we could...

Solve one equation for "a" and use the Substitution Method to solve the system:
Dividing the first equation by

which simplifies to:

Then substitute this into the other equation:

which, after adding e's exponents, simplifies to:

Divide by 1000:

etc.
Perhaps even faster...
Starting from:

and

Divide the two equations!

On the right side the a's cancel and, after subtracting e's exponents, we get:

etc.

I'll let you choose which solution, your corrected path or one of the two alternatives I've shown, to use to finish the problem.
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