Question

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

Answer 1

`\left( (27x^3)/(8y^9)\right)^ (5)/(3) \\\\\\ =\left( ((3x)^3)/((2y^3)^3)\right)^ (5)/(3) \\\\\\ = \frac{(3x)^{3 * (5)/(3) }}{(2y^3)^{3 * (5)/(3) }} \\\\\\ =((3x)^5)/((2y^3)^(5 )) \\\\\\ =(243x^5)/(32y^(15))`

Now, If the exponent was negative like you asked....


`\left( (27x^3)/(8y^9)\right)^ {-(5)/(3)} \\\\\\ =\left( (8y^9)/(27x^3)\right)^ {(5)/(3)}\\\\\\ =\left( ((2y^3)^3)/((3x)^3)\right)^ (5)/(3) \\\\\\ = \frac{(2y^3)^{3 * (5)/(3) }}{(3x)^{3 * (5)/(3) }} \\\\\\ =((2y^3)^(5 ))/((3x)^5) \\\\\\ =(32y^(15))/(243x^5)`