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

**cjh824** · September 22nd 2008, 05:03 PM

The question goes a little something like this, and sorry if it`s 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!

**skeeter** · September 22nd 2008, 05:34 PM

is this what you mean?

$\frac{\sqrt{x^3} \cdot \sqrt[5]{x^6}}{\sqrt[15]{x^3}}$

**cjh824** · September 22nd 2008, 05:59 PM

Yep! That's exactly what I mean!

**11rdc11** · September 22nd 2008, 06:19 PM

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

$x^{\frac{15}{6}}$

**Shyam** · September 22nd 2008, 06:36 PM

Originally Posted by cjh824

The question goes a little something like this, and sorry if it`s 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,

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

**11rdc11** · September 22nd 2008, 06:44 PM

Originally Posted by Shyam

further,

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

That should be

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

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