# Thread: Unable to solve roots expression

1. ok all I can do is to give examples with little explanation

any number you can write it in power of 1

$\displaystyle x=x^{1}$

$\displaystyle y^2=(y^2)^{1}$

$\displaystyle \frac{1}{x^2}=\frac{1}{(x^2)^{1}}$ so if I would to move the denominator to the numerator you will have

$\displaystyle \frac{1}{x^2}=\frac{1}{(x^2)^{1}}=(x^2)^{-1} = x^{2(-1)}=x^{-2}$

I think it is ok for now right....

$\displaystyle \sqrt[3]{x}=\sqrt[3]{x^{1}} = (x^{1})^{\frac{1}{3}}=x^{1(\frac{1}{3})}=x^{\frac{ 1}{3}}$

$\displaystyle \frac{1}{\sqrt[3]{x}} = \frac{1}{\sqrt[3]{x^{1}}}=\frac{1}{(x^{1})^{\frac{1}{3}}}=\frac{1}{ x^{1(\frac{1}{3})}}=\frac{1}{x^{\frac{1}{3}}}=x^{(-1)(\frac{1}{3})}=x^{\frac{-1}{3}}$

$\displaystyle \sqrt{xy} = (xy)^\frac{1}{2}$ this is clear right

$\displaystyle \frac{1}{\sqrt{xy}} = \frac{1}{(xy)^{\frac{1}{2}}} = ((xy)^{\frac{1}{2}})^{-1} = (xy)^{(-1)(\frac{1}{2})}=(xy)^{\frac{-1}{2}}$

$\displaystyle \frac{1}{\sqrt[3]{xy}}=\frac{1}{(xy)^{\frac{1}{3}}}=\left((xy)^{\fr ac{1}{3}}\right)^{-1} = (xy)^{(\frac{1}{3})(-1)}=(xy)^{\frac{-1}{3}}$

$\displaystyle \sqrt[5]{x^2y} = \sqrt[5]{(x^2y)^{1}} = \left((x^2y)^{1}\right)^{\frac{1}{5}} = (x^2y)^{(1)(\frac{1}{5})} = (x^2y)^{\frac{1}{5}}$

$\displaystyle \frac{1}{\sqrt[5]{x^2y}}=\frac{1}{\sqrt[5]{(x^2y)^{1}}}=$$\displaystyle \frac{1}{((x^2y)^{1})^{\frac{1}{5}}}=\frac{1}{(x^2 y)^{\frac{1}{5}}}=((x^2y)^{\frac{1}{5}})^{-1}=$$\displaystyle (x^2y)^{\left(\frac{1}{5}\right)(-1)}=(x^2y)^{\frac{-1}{5}}$

it is clear or not , is there anything else or anything is not clear ?

2. Thanks Amers the examples are helpful, but to make sure I got the hang of it I created several examples to see if I created them as exponents correctly.

$\displaystyle \frac{1}{\sqrt[5]{3xy^7}}= (3xy^7)^{-1/5}$

$\displaystyle \frac{1}{\sqrt[8]{2xyz^7}}=(2xyz^7)^{-1/8}$

$\displaystyle \sqrt[5]{x^4}=x^{4/5}$

$\displaystyle \sqrt[7]{7x^4}=(7x^4)^{1/7}$

$\displaystyle \sqrt[5]{(7x-4)^3}=(7x-4)^{3/5}$

Page 2 of 2 First 12