2(x + 1)^(3/2) = 27/4 ...........we deal with powers last, so begin by getting rid of the 2
=> (x + 1)^(3/2) = 27/8 ..........divided both sides by 2
now we want to get rid of the power. that is, we want the power of x + 1 to be 1. how do we do that? note that 3/2 * 2/3 = 1. so if we raise both sides to the 2/3 power, we make the left side less complicated.
=> [(x + 1)^(3/2)]^(2/3) = (27/8)^(2/3)
=> x + 1 = (27/8)^(2/3)
=> x = (27/8)^(2/3) - 1
how do we eveluate (27/8)^(2/3)?
remember, when we have a fractional power, think of "Flower Power"
you treat the fraction as a flower basically, yeah i know it's weird. the division sign represents the ground, the numerator on top represent the flower (which rhymes with power, hence flower-power), and the denominator under the division sign represents the root. so in short, when we have a fractional power, the numerator represents the power, and the denominator represents the root. so,
x = (27/8)^(2/3) - 1
=> x = cuberoot[(27/8)^2]- 1
we can find the root or the power first, it is easier to find the root:
=> x = (3/2)^2 - 1
=> x = 9/4 - 1
=> x = 5/4
you could also split the powers up:
(27/8)^(2/3) = (27/8)^[2 * 1/3] = [(27/8)^2]^(1/3) = [(27/8)^(1/3)]^2
note that we can take either power first, i found it easier to take the 1/3 power first, which is the same as the cuberoot
so (27/8)^(2/3) = [(27/8)^(1/3)]^2 = (3/2)^2 = 9/4