∫e^(-2x) dx
Let u = 2x
du= 2 dx
dx=(1/2) du
∫e^(-2x) dx = (1/2) ∫e^-u du
= (1/2) (-e^-u)
= -e^-u /2 + C
= -e^-(2x) / 2 + C
Chat with our AI personalities
e 2x = (1/2) e 2x + C ============
The power law of indices says: (x^a)^b = x^(ab) = x^(ba) = (x^b)^a → e^(2x) = (e^x)² but e^x = 2 → e^(2x) = (e^x)² = 2² = 4
The definition of the natural log ln of a number is the power that you have to raise e to in order to get that number. Therefore, ln(2x+3) is the power you have to raise e to to get 2x + 3.
-2y square exp power -2x-1
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)