Rewrite the rational exponent as a radical. 5 to the 3 over 4 power, to the 2 over 3 power Answer Choices: A) the cube root of 5 squared B) the twelfth root of 5 C) the square root of 5 D)the cube root of 5 the fourth power I think it's A but im not sure, someone please help?
\[\huge\rm (5^\frac{ 5 }{ 4 })^\frac{ 2 }{ 3 }\] like this ?
@Nnesha Yes.
well can you take a screenshot i think i typed it wrong
@Nnesha
@Nnesha did you get it?
weird question .... why do you think it's A ?
@Nnesha When I worked out the problem I some how ended up with A
well i didn't get A i got*NONE* ORR i'm not paying attention
@freckles ???
you have to apply exponent rule \[\huge\rm (x^m)^n = x^{m \times n}\] multiply exponents and then reduce the fraction
@Nnesha I did so, I did it 4 more times and got different answers each time. I have no idea anymore.
hey
hey @Nnesha he said to ...to the 3/4 power... but then he said yes when you put 5/4 instead...
\[(5^{\frac{3}{4})\frac{2}{3}} =5^{\frac{3}{4} \cdot \frac{2}{3}}\] first look at the fraction multiplication action going on there: \[\frac{3}{4} \cdot \frac{2}{3}\] can you simplify that?
basically you need to simplify 3/3 and 2/4
*facepalm* it's 3/4 >.<
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\huge\rm (5^\frac{ 5 }{ 4 })^\frac{ 2 }{ 3 }\] like this ? \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @06292014 @Nnesha Yes. \(\color{blue}{\text{End of Quote}}\) ......
wow my mistake guys :/ im so sorry
my bad.. thanks freckles
does @06292014 still need help? if so have you done the multiplication I asked about?
@freckles I got 1/2
cool so you have 5^(1/2)
which is the same as square root of 5
@freckles @Nnesha thank you so much both of you for your help!
@Nnesha @freckles
look at the exponents do top exponent - bottom exponent
i can't zoomin in this pdf http://assets.openstudy.com/updates/attachments/557890d9e4b0636b8cc2974f-06292014-1433965330041-cuserspattydocumentsscreenshot.pdf and people like me barely can see this^ ;P
and write these down 'n ur notes \[\huge\rm x^m \times x^n = x^{m+n}\]\[\large\rm (x^m)^n =x^{m \times n}\]\[\large\rm \frac{ x^m }{ x^n }= x^{m-n}\]
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