SOLUTION: Can't find any lesson that explains solving equations containing a rational exponent on the variable. Here's my problem: (there are lessons for more in-depth problems, but I need

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Can't find any lesson that explains solving equations containing a rational exponent on the variable. Here's my problem: (there are lessons for more in-depth problems, but I need       Log On


   

Question 288386: Can't find any lesson that explains solving equations containing a rational exponent on the variable.
Here's my problem: (there are lessons for more in-depth problems, but I need to learn the simple stuff before moving on to the more complicated stuff)
%28x%29%5E%284%2F3%29=81 (x to the 4/3 power=81)
I used
%28x%29%5E%284%2F3%29=81
%282x%29%5E%281%2F3%29=81
%282x%29%5E%282%2F3%29%29%5E%281%2F3%29=81%5E%282%2F3%29
which made
3x=%2881%29%5E%282%2F3%29
and then divided by 3, so
x=%2827%29%5E%281%2F3%29
Does this look right?
I didn't think I was supposed to have any power left afterwards.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%29%5E%284%2F3%29=81 (x to the 4/3 power=81)
I used
%28x%29%5E%284%2F3%29=81
%282x%29%5E%281%2F3%29=81 **** this is not good
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x%5E%281%2F3%29+=+3+ is better
Then x = 27
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Also, x = -27
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There are other solutions with complex numbers, but this is proably sufficient for now.
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%282x%29%5E%282%2F3%29%29%5E%281%2F3%29=81%5E%282%2F3%29
which made
3x=%2881%29%5E%282%2F3%29
and then divided by 3, so
x=%2827%29%5E%281%2F3%29
Does this look right?