Question 718152
{{{y = a*e^(-bx)}}} when x=0.6, y=1000 and when x=1.8, y=1<br>
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.<br>
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.<br>
Your work is find up to:
{{{-bx = log(e, (y/a))}}}
But to "move x over" you have to divide by x, not multiply it:
{{{-b = log(e, (y/a))/x}}}
Substituting in the values for x and y we get two equations:
{{{-b = log(e, (1000/a))/0.6}}}
and
{{{-b = log(e, (1/a))/1.8}}}
Using the Substitution Method on this system, we would substitute one equation's expression for -b in for the other equation's -b:
{{{log(e, (1000/a))/0.6  = log(e, (1/a))/1.8}}}
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.<br>
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:
{{{y = a*e^(-bx)}}}
giving us:
{{{1000 = a*e^(-b(0.6))}}}
and
{{{1 = a*e^(-b(1.8))}}}
which simplify to:
{{{1000 = a*e^(-0.6b)}}}
and
{{{1 = a*e^(-1.8b)}}}
From here we could...<ul><li>Solve one equation for "a" and use the Substitution Method to solve the system:
Dividing the first equation by {{{e^(-0.6b)}}}
{{{1000/e^(-0.6b) = a}}}
which simplifies to:
{{{1000e^(0.6b) = a}}}
Then substitute this into the other equation:
{{{1 = (1000e^(0.6b))e^(-1.8b)}}}
which, after adding e's exponents, simplifies to:
{{{1 = 1000e^(-1.2b)}}}
Divide by 1000:
{{{0.001 = e^(-1.2b)}}}
{{{log(e, (0.001)) = log(e, (e^(-1.2b)))}}}
{{{log(e, (0.001)) = (-1.2b)*log(e, (e))}}}
{{{log(e, (0.001)) = -1.2b}}}
{{{log(e, (0.001))/(-1.2) = b}}}
etc.</li><li>Perhaps even faster...
Starting from:
{{{1000 = a*e^(-0.6b)}}}
and
{{{1 = a*e^(-1.8b)}}}
Divide the two equations!
{{{1000/1 = (a*e^(-0.6b))/(a*e^(-1.8b))}}}
On the right side the a's cancel and, after subtracting e's exponents, we get:
{{{1000 = e^(1.2b)}}}
etc.</li></ul>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.