Negative radical exponent simplification

**alex95** · January 6th 2011, 11:28 AM

Hi,

I am dealing with the following problem and am unsure whether
I solved it right:
=================================================
(16xy)^(1/4) (-8x^2z^3)^(-1/3)
--------------------------------
(25^-1 x^2 yz)^(-1/2)

2(xy)^(1/4) (-8x^2z^3)^(-1/3)
= ------------------------------
((25^-1 x^2 yz)^-1)^(1/2))

2(xy)^(1/4) (-8x^2z^3)^(-1/3)
= --------------------------------
(25x^-2y^-1z^-1)^(1/2)

2(xy)^(1/4)
= ---------------------------------------------
5(x^-2y^-1z^-1)^(1/2) (-8x^2z^3)^(1/3)

= 2(xy)^(1/4)
---------------------------------
5(x^-2y^-1z^-1)^(1/2) -2z(x^2)^(1/3)
===================================

Thanks for the help,

Alex

**dwsmith** · January 6th 2011, 12:18 PM

$\displaystyle\frac{(16xy)^{1/4}*(-8x^2z^3)^{-1/3}}{(25^{-1}x^2yz)^{-1/2}}$

Is this is the expression?

**alex95** · January 6th 2011, 12:24 PM

Yes sir

**alex95** · January 6th 2011, 12:25 PM

Actually, the 1/2 exponent in the denominator is negative 1/2 (-1/2)

**pickslides** · January 6th 2011, 12:43 PM

It did it on the fly, by I get

$\displaystyle -x^{\frac{7}{12}}y^{\frac{3}{4}}z^{\frac{-2}{3}}$

**alex95** · January 6th 2011, 12:58 PM

Would you mind showing the steps for my learning purposes?

Thanks,

Alex

**pickslides** · January 6th 2011, 01:11 PM

Originally Posted by alex95

Would you mind showing the steps for my learning purposes?

No problem. Where possible for all terms

Step 1: apply $\displaystyle (a^m)^n = a^{m\times n}$

Step 2: apply $\displaystyle a^m\times a^n = a^{m+ n}$

Step 3: apply $\displaystyle a^m \div a^n = a^{m- n}$

**Soroban** · January 6th 2011, 01:15 PM

Hello, alex95!

$\displaystyle\text{SimpliftyL } \;\frac{(16xy)^{\frac{1}{4}}(-8x^2z^3)^{-\frac{1}{3}}} {(2s^{-1}x^2yz)^{-\frac{1}{2}}}$

We have: . $\displaystyle \frac{(16xy)^{\frac{1}{4}}(2s^{-1}x^2yz)^{\frac{1}{2}}} {(-8x^2z^3)^{\frac{1}{3}}}$

. . . . . . $\displaystyle =\; \frac{16^{\frac{1}{4}}\cdot x^{\frac{1}{4}}\cdot y^{\frac{1}{4}} \cdot 2^{\frac{1}{2}}\cdot(s^{-1})^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}\cdot y^{\frac{1}{2}}\cdot z^{\frac{1}{2}}} {(-8)^{\frac{1}{3}}(x^2)^{\frac{1}{3}}(z^3)^{\frac{1} {3}}}$

. . . . . . $\displaystyle =\;\frac{2\cdot x^{\frac{1}{4}}\cdot y^{\frac{1}{4}}\cdot 2^{\frac{1}{2}}\cdot x\cdot y^{\frac{1}{2}}\cdot z^{\frac{1}{2}}\cdot s^{-\frac{1}{2}}} {-2\cdot x^{\frac{2}{3}}\cdot z}$

. . . . . . $\displaystyle =\; \frac{2^{\frac{3}{2}}\cdot x^{\frac{5}{4}} \cdot y^{\frac{3}{4}}\cdot z^{\frac{1}{2}}\cdot s^{-\frac{1}{2}}} {-2\cdot x^{\frac{2}{3}}\cdot z}$

. . . . . . $\displaystyle\;=-\frac{2^{\frac{1}{2}}\cdot x^{\frac{7}{12}}\cdot y^{\frac{3}{4}}} {z^{\frac{1}{2}}\cdot s^{\frac{1}{2}}}$

**Archie Meade** · January 6th 2011, 01:16 PM

$\displaystyle\frac{(16xy)^{\frac{1}{4}}\;\left(-8x^2z^3\right)^{-\frac{1}{3}}}{\left(25^{-1}x^2yz\right)^{-\frac{1}{2}}}=\frac{16^{\frac{1}{4}}x^{\frac{1}{4} }y^{\frac{1}{4}}\;\left(25^{-1}x^2yz\right)^{\frac{1}{2}}}{\left(-8x^2z^3\right)^{\frac{1}{3}}}$

$=\displaystyle\frac{2x^{\frac{1}{4}}y^{\frac{1}{4} }xy^{\frac{1}{2}}z^{\frac{1}{2}}}{-2zx^{\frac{2}{3}}25^{\frac{1}{2}}}=\frac{x^{\frac{ 1}{4}}x^1y^{\frac{1}{4}}y^{\frac{1}{2}}z^{\frac{1} {2}}}{-5x^{\frac{2}{3}}z}$

$=\displaystyle\ -\frac{x^{\left(\frac{5}{4}-\frac{2}{3}\right)}y^{\frac{3}{4}}z^{\left(\frac{1 }{2}-1\right)}}{5}$

$=-\displaystyle\frac{x^{\frac{7}{12}}y^{\frac{3}{4}} }{5z^{\frac{1}{2}}}$

**pickslides** · January 6th 2011, 01:20 PM

Originally Posted by Soroban

$\displaystyle\;=-\frac{2^{\frac{1}{2}}\cdot x^{\frac{7}{12}}\cdot y^{\frac{3}{4}}} {z^{\frac{1}{2}}\cdot s^{\frac{1}{2}}}$

Another great post Soroban, but I don't see where you found an 's' in all of that... I think Archie has it nailed.

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