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 question is ambiguous and two possible answers are given below: 2x2 - x2 = x2 or (2x)2 - x2 = 4x2 - x2 = 3x2
∫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
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)
If that was 2x^2 + 5x + 2, it would factor to (2x + 1)(x + 2)
-2x-2x-2x-2x-2=-32
-2x-2x-2x-2= 16 (positive)
-2 to the 4th power is 16.
3x^2 + x + 2x^2 = 5x^2 + x
2(x + 2)(x - 2)(x2 + 2x + 4)(x2 - 2x + 4)
-2 to the power of two is the same as -2x-2 Two negative numbers multiplied gives you a positive number. So, -2x-2=4 -2 to the power of 2=4
2x^2
2 with a power of 2x 5 with a power of 2
(x^2 - 2x + 2)(x^2 + 2x + 2)
(2x - 9)(2x - 9) or (2x - 9)2
d/dx(2x) = 2 simple power rule
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