If \(\frac{(a^2 b^{-3}c)^{\frac{3}{4}}}{a^{-1}b^{4}c^{5}}=a^{p} b^{q} c^{r}\) What is the value of p+2q?

Explanation

Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at

p = \(\frac{5}{2}\), q = -\(\frac{25}{4}\) and r = -\(\frac{17}{4}\)

NUMERATOR:   a\(^{\frac{2}{1}}*{\frac{3}{4}}\)          b\(^{\frac{-3}{1}}*{\frac{3}{4}}\)          c\(^{\frac{1}{1}}*{\frac{3}{4}}\)

                           a\(\frac{3}{2}\)      b\(\frac{-9}{4}\)    c\(\frac{3}{4}\)

Using index method for numerator & denominator :  a\(^{\frac{3}{2}}-{\frac{-1}{1}}\)           b\(^{\frac{-9}{4}}-{\frac{4}{1}}\)           c\(^{\frac{3}{4}}-{\frac{5}{1}}\)

:  a\(\frac{5}{2}\)      b\(\frac{-25}{4}\)    c\(\frac{-17}{4}\)

Then p+2q will give you \(\frac{5}{2}+2\left(\frac{-25}{4}\right)= -10\)