How do you simplify:
5*e^(-10x) - 3*e^(-20x) = 2
I'm not sure if I can take natural log of both sides to bring down the exponents.
I will assume you want to "solve" for x
let e^(-10x) = y , then e^(-20x) = (e^(-10x)^2 = y^2
then you have
5y - 3y^2 = 2
3y^2 - 5y + 2 = 0
(y-1)(3y-2) = 0
y = 1 or y = 2/3
so e^(-10x) = 1 or e^(-10x) = 2/3
if e^(-10x) = 1
ln e^(-10x) = ln 1 = 0
-10x lne = 0
-10x = 0
x = 0
if e^(-10x) = 2/3
...
-10x = ln2 - ln3
x = -(l2 - ln3)/10 = appr .04055