SOLUTION: (8x^(1/2)y^3)^(1/3)•(25xy^(1/2))^(1/2)

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: (8x^(1/2)y^3)^(1/3)•(25xy^(1/2))^(1/2)      Log On


   

Question 1154462: (8x^(1/2)y^3)^(1/3)•(25xy^(1/2))^(1/2)
Found 2 solutions by Edwin McCravy, MathLover1:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

%288x%5E%281%2F2%29y%5E3%29%5E%281%2F3%29%2A%2825xy%5E%281%2F2%29%29%5E%281%2F2%29

Put in all the invisible multiplication symbols:



Put in all invisible 1-exponents, to the 8, the 25 and the x:



Remove the parentheses by distributing the exponents:



Add the exponents of the x's and of the y's:









Since
      matrix%282%2C1%2C%22%22%2C8%5E%281%2F3%29=root%283%2C8%29=2%29
and 
      matrix%282%2C1%2C%22%22%2C25%5E%281%2F2%29=sqrt%2825%29=5%29
,

matrix%282%2C1%2C%22%22%2C2%2Ax%5E%282%2F3%29%2Ay%5E%285%2F4%29%2A5%29

matrix%282%2C1%2C%22%22%2C10%2Ax%5E%282%2F3%29%2Ay%5E%285%2F4%29%29

If your teacher wanted you to stop with fractional exponents, then
the above would be the final answer.  However, if your teacher
wanted your answer to be expressed with roots, then continue:

Express the fractional exponents with the same LCD:

matrix%282%2C1%2C%22%22%2C10%2Ax%5E%288%2F12%29%2Ay%5E%2815%2F12%29%29

Since
      matrix%282%2C1%2C%22%22%2Cx%5E%288%2F12%29=root%2812%2Cx%5E8%29%29
and 
      matrix%282%2C1%2C%22%22%2Cy%5E%2815%2F12%29=root%2812%2Cy%5E15%29%29
      
matrix%282%2C1%2C%22%22%2C10%2Aroot%2812%2Cx%5E8%29%2Aroot%2812%2Cy%5E15%29%29

matrix%282%2C1%2C%22%22%2C10%2Aroot%2812%2Cx%5E8%2Ay%5E15%29%29

Edwin

Answer by MathLover1(20850) About Me  (Show Source):