H hydride Sep 2009 51 0 May 23, 2010 #1 \(\displaystyle (\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{-4} = \) First i would change it too: \(\displaystyle \frac{1}{(\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{4}} = \) Is that correct? I need some more guidance.(Doh) Thanks.

\(\displaystyle (\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{-4} = \) First i would change it too: \(\displaystyle \frac{1}{(\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{4}} = \) Is that correct? I need some more guidance.(Doh) Thanks.

S SpringFan25 May 2010 1,034 272 May 23, 2010 #2 its a sensible first step, i'd simplify the x's inside the bracket next \(\displaystyle \frac{x^{\frac{-2}{3}}}{x^{0.5}} = x^{\frac{-2}{3}-0.5} = x^{\frac{-7}{6}}\)

its a sensible first step, i'd simplify the x's inside the bracket next \(\displaystyle \frac{x^{\frac{-2}{3}}}{x^{0.5}} = x^{\frac{-2}{3}-0.5} = x^{\frac{-7}{6}}\)

H hydride Sep 2009 51 0 May 23, 2010 #3 SpringFan25 said: its a sensible first step, i'd simplify the x's inside the bracket next \(\displaystyle \frac{x^{\frac{-2}{3}}}{x^{0.5}} = x^{\frac{-2}{3}-0.5} = x^{\frac{-7}{6}}\) Click to expand... Oh okay thanks. so \(\displaystyle \frac{1}{(\frac{y^{\frac{3}{4}}}{x^{\frac{-7}{6}}})^4} \) Whats next?

SpringFan25 said: its a sensible first step, i'd simplify the x's inside the bracket next \(\displaystyle \frac{x^{\frac{-2}{3}}}{x^{0.5}} = x^{\frac{-2}{3}-0.5} = x^{\frac{-7}{6}}\) Click to expand... Oh okay thanks. so \(\displaystyle \frac{1}{(\frac{y^{\frac{3}{4}}}{x^{\frac{-7}{6}}})^4} \) Whats next?

B bjhopper Nov 2007 985 175 Trumbull Ct May 23, 2010 #4 hydride said: \(\displaystyle (\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{-4} = \) First i would change it too: \(\displaystyle \frac{1}{(\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{4}} = \) Is that correct? I need some more guidance.(Doh) Thanks. Click to expand... in Hi hydride, multiply exponents and simplify each term of the numerator and denominator must be raised to -4.answer will be of in termof x to a power andy to a power example (x^2)^1/2=x 2x1/2=1 thats 2 times 1/2=1 bjh Last edited: May 23, 2010

hydride said: \(\displaystyle (\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{-4} = \) First i would change it too: \(\displaystyle \frac{1}{(\frac{x^{-\frac{2}{3}}y^\frac{3}{4}}{x^\frac{1}{2}})^{4}} = \) Is that correct? I need some more guidance.(Doh) Thanks. Click to expand... in Hi hydride, multiply exponents and simplify each term of the numerator and denominator must be raised to -4.answer will be of in termof x to a power andy to a power example (x^2)^1/2=x 2x1/2=1 thats 2 times 1/2=1 bjh