radical notation

[claudette]

does anyone know how to write 16^-3/4 in radical notation? im stuck on this

 

[Deleted member 4993]

does anyone know how to write 16^-3/4 in radical notation? im stuck on this

I'll do a different but similar problem for you:

$\displaystyle (125)^{\frac{-2}{3}} \ = \ \dfrac{1}{\sqrt[3]{125^2}}$

 

[claudette]

I'll do a different but similar problem for you:

$\displaystyle (125)^{\frac{-2}{3}} \ = \ \dfrac{1}{\sqrt[3]{125^2}}$

i actually figured that part out now im trying to figure out how to evaluate the expression now

 

[Deleted member 4993]

i actually figured that part out now im trying to figure out how to evaluate the expression now

Okay ... go at it ... show your work if you need help.

 

[HallsofIvy]

If you consider the factors of 16, the fourth root should be clear.

 

[Deleted member 4993]

If you give the factors of 125 some thought, the cube root should be clear.

Her problem is to evaluate $\displaystyle \displaystyle \left [16\right ]^{\frac{-3}{4}}$.

The problem with 125 was made up by me.

 

[lookagain]

does anyone know how to write 16^-3/4 in radical notation? im stuck on this

Claudette, you must use grouping symbols, such as in "16^(-3/4)." Also, for appropriate numbers, as in this case, the expression is equivalent to: $\displaystyle \ \ \dfrac{1}{(\sqrt[4]{16})^3}$ $\displaystyle \ \ \ \ \ Continue \ \ with \ \ that.$