Respuesta :
x^a/b is [tex] \sqrt[b]{x^a} [/tex] . The way I memorise that is x^1/3 is the cubic root of x. Do you get it? In that case, x is raised to a power of 1 and the cubic root is practically has a power of 3.
In your example,
[tex] \sqrt[ \frac{3}{2} ]{16 x^4}[/tex] is practically square rooting each term then cubing them individually. Remember when square-rooting any index you halve it. I'll elaborate:
[tex] \sqrt{x^4} [/tex] = [tex] x^{2} [/tex]
[tex] \sqrt{16} [/tex] = 4
Then cube each,
[tex]4^3[/tex] = 64
and [tex]( x^{2} )^3[/tex] = [tex] x^{6} [/tex]
As for the 2nd part: you must use the rules of indices.
[tex] x^{a} * x^{b} = x^{a+b} [/tex]
So breaking the question up:
3 * 3 = 9
[tex] x^{ \frac{1}{2} } [/tex] stays as is since the 2nd term does not contain x
now:
[tex] y^{ \frac{4}{3} } * y^{1} = y^{ \frac{4}{3} + 1 } = y^{ \frac{4}{3} + \frac{3}{3} } = y^{ \frac{7}{3} } [/tex]
This makes your final answer look like this:
[tex]9 x^{ \frac{1}{2} } y^{ \frac{7}{3} } [/tex]
I hope that helped and good luck in your test!
In your example,
[tex] \sqrt[ \frac{3}{2} ]{16 x^4}[/tex] is practically square rooting each term then cubing them individually. Remember when square-rooting any index you halve it. I'll elaborate:
[tex] \sqrt{x^4} [/tex] = [tex] x^{2} [/tex]
[tex] \sqrt{16} [/tex] = 4
Then cube each,
[tex]4^3[/tex] = 64
and [tex]( x^{2} )^3[/tex] = [tex] x^{6} [/tex]
As for the 2nd part: you must use the rules of indices.
[tex] x^{a} * x^{b} = x^{a+b} [/tex]
So breaking the question up:
3 * 3 = 9
[tex] x^{ \frac{1}{2} } [/tex] stays as is since the 2nd term does not contain x
now:
[tex] y^{ \frac{4}{3} } * y^{1} = y^{ \frac{4}{3} + 1 } = y^{ \frac{4}{3} + \frac{3}{3} } = y^{ \frac{7}{3} } [/tex]
This makes your final answer look like this:
[tex]9 x^{ \frac{1}{2} } y^{ \frac{7}{3} } [/tex]
I hope that helped and good luck in your test!
