Question:`((2^(1/2) xx 3^(1/3) xx 4^(1/4))/(10^(- 1/5) xx 5^(3/5))// (3^(4/3) xx 5^(- 7/5))/(4^(- 3/5) xx 6))`
Answer
`(2^(1/2) xx 3^(1/3) xx 4^(1/4))/(10^(- 1/5) xx 5^(3/5)) xx (4^(- 3/5) xx 6)/(3^(4/3) xx 5^(- 7/5))` = `(2^(1/2) xx 3^(1/3) xx 2^(1/2))/(2^(- 1/5) xx 5^(- 1/5) xx 5^(3/5)) xx ((2^2)^(- 3/5) xx (2 xx 3))/(3^(4/3) xx 5^(- 7/5))` = `(2^(1/2) xx 3^(1/3) xx 2^(1/2))/(2^(- 1/5) xx 5^(- 1/5) xx 5^(3/5)) xx (2^(- 6/5) xx 2 xx 3)/(3^(4/3) xx 5^(- 7/5))` = `(2^(1/2 + 1/2 +1/5 - 6/5 +1) . 3^(1/3 +1 - 4/3))/5^(-1/5 + 3/5 - 7/5)` = `(2^1 xx 3^0 xx 5^1)` = (2 x 1 x 5) = 10