# Question with roots.indexes, etc. So confusing for homework pass-in.

• Sep 22nd 2008, 06:03 PM
cjh824
Question with roots.indexes, etc. So confusing for homework pass-in.
The question goes a little something like this, and sorry if its confusing:

(Square root of x^3)times (index of 5, square root of x^6)

all divided by (index of 15, square root of x^3)

Any help would be so greatly appreciated. Thanks a million!
• Sep 22nd 2008, 06:34 PM
skeeter
is this what you mean?

$\frac{\sqrt{x^3} \cdot \sqrt[5]{x^6}}{\sqrt[15]{x^3}}$
• Sep 22nd 2008, 06:59 PM
cjh824
Yep! That's exactly what I mean!
• Sep 22nd 2008, 07:19 PM
11rdc11
Think of it like this

$\frac{(x^{\frac{3}{2}})(x^{\frac{6}{5}})}{x^{\frac {3}{15}}}$

add the exponents in the numerator since it to the same base

$\frac{x^{\frac{27}{10}}}{x^{\frac{3}{15}}}$

then bring the power in the denominator to the top

and you end up with

$x^{\frac{15}{6}}$
• Sep 22nd 2008, 07:36 PM
Shyam
Quote:

Originally Posted by cjh824
The question goes a little something like this, and sorry if its confusing:

(Square root of x^3)times (index of 5, square root of x^6)

all divided by (index of 15, square root of x^3)

Any help would be so greatly appreciated. Thanks a million!

further,

$= x^{\frac{5}{3}}=x \sqrt[3] {x^2}$
• Sep 22nd 2008, 07:44 PM
11rdc11
Quote:

Originally Posted by Shyam
further,

$= x^{\frac{5}{3}}=x \sqrt[3] {x^2}$

That should be

$= x^{\frac{5}{2}}= x^2\sqrt{x}$