Radical Powers and Radicals: n^t (2/3)=4

Missy

New member
Joined
Jan 16, 2008
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I hope that I am writing this right now sure how.
But the problem is n^t (2/3)=4
I want to know if I done this right someone showed me this way but it looks totally funny from the way its suppose to look
t=(4)3/2
t=((2cubrt))
2*2*2 =t((8))
t=8

Can someone please tell me if this is correct, with this problem I am more confuse now that I was before. Please help
 

Subhotosh Khan

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Jun 18, 2007
Messages
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Re: Radical Powers and Radicals

Missy said:
I hope that I am writing this right now sure how.
But the problem is n^t (2/3)=4<<< You don't really tell what the problem is. Find n? Find t? simplify? what?
I want to know if I done this right someone showed me this way but it looks totally funny from the way its suppose to look
t=(4)3/2
t=((2cubrt))<<< If 'cubrt' means 'cube root' - then this step is incorrect
2*2*2 =t((8))
t=8<<< answer is correct - assuming corrected problem statement below - but your steps should look as follows:
If the problem is:

Find 't' when

\(\displaystyle t^{\frac {2}{3}} \, = \, 4\)

\(\displaystyle t^{\frac {2}{3}} \, = \, 2^2\)

\(\displaystyle t^{(\frac {2}{3}\cdot\frac{1}{2})} \, = \, 2^{(2\cdot\frac{1}{2})}\)

\(\displaystyle t^{\frac {1}{3}} \, = \, 2^1 \, = \, 2\)

\(\displaystyle t^{(\frac {1}{3}\cdot 3)} \, = \, 2^{3}\)

\(\displaystyle t \, = \, 8\)
 
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