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# Plato Classroom Question #13

-1
744
4
+108

Match each expression to its equivalent expression with the rational denominators.

Tiles

$$^1/_{\sqrt[4]{3x^3y^5}}$$

$$^3/_{\sqrt[4]{27x^{11}y^{13}}}$$

$$^2/_{\sqrt[6]{2x^7y^5}}$$

$$^4/_{\sqrt[6]{32x^5y^9}}$$

Pairs

$$^{2\sqrt[6]{2xy^3}}/_{xy^2}\to$$

$$^{\sqrt[4]{27xy^3}}/_{3xy^2}\to$$

$$^{\sqrt[4]{3xy^3}}/_{x^3y^4}\to$$

$$^{\sqrt[6]{32x^5y}}/_{x^2y}\to$$

Jan 19, 2018

#1
+108675
+2

Hi SpaceModo,

I'll do one of them.

$$\frac{{2\sqrt[6]{2xy^3}}}{_{xy^2}}\\ =\frac{{2\sqrt[6]{2xy^3}}}{\sqrt[6]{x^6y^{12}}}\\ =\frac{{2\sqrt[6]{2}}}{\sqrt[6]{x^5y^{9}}}\times\frac{2^{5/6}}{2^{5/6}}\\ =\frac{{2*2}}{\sqrt[6]{x^5y^{9}}}\times\frac{1}{(2^5)^{1/6}}\\ =\frac{{4}}{\sqrt[6]{32x^5y^{9}}}\\$$

That is one done, now you can copy the technique and try matching up the others.

If you do not understand this one just ask and see if you can specify what line is giving you trouble :)

Jan 20, 2018
#4
+108
0

\begin{align*}...
{^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\sqrt[6]{2xy^3}}/_{xy^2}}\\
{^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\sqrt[6]{2xy^3}}/_{\sqrt[6]{32x^5y^9}}}\\
{^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\sqrt[6]{2}}/_{\sqrt[6]{x^5y^9}}}\times{^{2^{^{^5}/_6}}/_{2^{^5/_6}}}\\
{^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\times2}/_{\sqrt[6]{x^5y^9}}}\times{^1/_{(2^5)^{^1/_6}}}\\
{^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{\checkmark}{=}{^4/_{\sqrt[6]{32x^5y^9}}}
\end{align*}

\begin{align*} {^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\sqrt[6]{2xy^3}}/_{xy^2}}\\ {^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\sqrt[6]{2xy^3}}/_{\sqrt[6]{32x^5y^9}}}\\ {^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\sqrt[6]{2}}/_{\sqrt[6]{x^5y^9}}}\times{^{2^{^5/_6}}/_{2^{^5/_6}}}\\ {^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{?}{=}{^{2\times2}/_{\sqrt[6]{x^5y^9}}}\times{^1/_{(2^5)^{^1/_6}}}\\ {^4/_{\sqrt[6]{32x^5y^9}}}&\stackrel{\checkmark}{=}{^4/_{\sqrt[6]{32x^5y^9}}} \end{align*}

][   |) ()   \/ ≡ R y   ( () |\/| P |_ ≡ ><   |\/| /-\ T |-|   ( () |) | |\| G!

However, thank you very much for the help I needed!

SpaceModo  Jan 22, 2018
#2
+109345
+1

4√[ 27xy^3] / [ 3xy^2]      writing this in an exponential fashion, we have

[ (3^3)^(1/4) * x^(1/4)  * y^(3/4) ]  /  [  3xy^2]  =

[( 3^3)(1/4)  * x^(1/4) * y^(3/4)  /  [  (3^(4/4) * x^(4/4) * y^(8/4)  ]

[ 3^(3/4)  * x^(1/4)  * y^(3/4) ]  /  [  3^(4/4) * x^(4/4) * y^(8/4)  ]  =

{ Using   a^m / a^n  =   a^(m - n)  }

1  /  [ 3^(1/4) * x^(3/4) * y^(5/4) ]   write back in radical form

1 / 4√ [ 3x3y5 ]

Jan 20, 2018
#3
+109345
+1

4√[ 3xy^3 ]  /  [ x^3y^4 ]  =

[  3^(1/4) * x^(1/4)  * y^(3/4) ] / [ x^(12/4) *y^(16/4) ]  =

3^(1/4)  / [  x^(11/4) * y^(13/4) ]  =

Multiply  top/bottom  by  3^(3/4)  =

[ 3^(1/4) * 3^(3/4)] /  [ x^(11/4) * y^(13/4)  * 3^(3/4) ]  =

3 / [  x^(11/4) * y^(13/4) * 27^(1/4)  ]

Write back in radical form

3 / 4√ [  27  x11 y13 ]

Jan 20, 2018