How do you solve (2x-5)^(3/2) = 27?

Question

How do you solve (2x-5)^(3/2) = 27?

Answer 1

The difficulty in this problem is in dealing with the exponent (3/2). So, think of some clever way to get the exponent to be easier---say 1.  Well, if you raise an exponent to a power, the new exponent is the product of the old times the clever power

(a^m)^n = a^(mn)

So, if you raise (2x - 5)^(3/2) to the 2/3 power, it becomes

[(2x - 5)^(3/2)]^(2/3) = (2x -5)^(3/2 x 2/3) = 2x - 5

But since we raised the left side of the equation to the power (2/3), we must do the same to the right side.  Therefore,

27^(2/3) means the cube root of 27-squared, or 9

Therefore, 2x - 5 = 9

x = 7

Check.

Answer 2

sqrt [ (2x-5)^3 ] = 27

(2x-5)^3 = (3^3)^2

(2x-5)^3 = (3^2)^3

2x-5 = 3^2

2x-5 = 9

2x= 14

x=7