1. ## Negative radical exponent simplification

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

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

Is this is the expression?

3. Yes sir

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

5. It did it on the fly, by I get

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

6. Would you mind showing the steps for my learning purposes?

Thanks,

Alex

7. 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 $\displaystyle a^m \div a^n = a^{m- n}$

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

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

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

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

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

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

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

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

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

10. 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.