• Mar 20th 2008, 05:32 AM
Hanga
Okay so I've been doing alot of work with exponentials and now i've gone stuck on a leap from using simple exponents to using several.

Here is what I could really need to have explained to me;

(16x^6)^2/3

So im thinking something like taking the third square of 16x^6, however I don't know if I can take the square out of something that has an exponential.
I'd appreciate anyone sheding some light on this for me :)
• Mar 20th 2008, 05:51 AM
colby2152
Quote:

Originally Posted by Hanga
Okay so I've been doing alot of work with exponentials and now i've gone stuck on a leap from using simple exponents to using several.

Here is what I could really need to have explained to me;

(16x^6)^2/3

So im thinking something like taking the third square of 16x^6, however I don't know if I can take the square out of something that has an exponential.
I'd appreciate anyone sheding some light on this for me :)

$\displaystyle (16x^6)^{2/3}$

$\displaystyle \left[(16x^6)^2\right]^{1/3}$

$\displaystyle (256x^{12})^{1/3}$

$\displaystyle 256^{1/3}x^4$

$\displaystyle \left[4^{4}\right]^{1/3}x^4$

$\displaystyle 4^{4/3}x^4$

$\displaystyle 4*4^{1/3}x^4$

$\displaystyle 4x^4 .^3\sqrt{4}$
• Mar 20th 2008, 06:03 AM
teuthid
Relevant Exponent laws
There are two exponent laws that you need to employ:

(xy)^p = (x^p)(y^P)
(x^p)^q = x^(pq)

These laws work for any real exponents (even if they're negative or fractions). I've attached a document that shows how these laws are applied in your particular example.

Another key idea involved with this problem is that of fractional exponents' denominators representing roots. In your example, the 2/3 exponent is telling you to square the cubed root of the expression.
• Mar 20th 2008, 07:53 AM
Hanga
Quote:

Originally Posted by teuthid
There are two exponent laws that you need to employ:

(xy)^p = (x^p)(y^P)
(x^p)^q = x^(pq)

These laws work for any real exponents (even if they're negative or fractions). I've attached a document that shows how these laws are applied in your particular example.

Another key idea involved with this problem is that of fractional exponents' denominators representing roots. In your example, the 2/3 exponent is telling you to square the cubed root of the expression.

Whaw! Thanks alot :)