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?
The correct answer is D. -10
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 \(\frac{5}{2}+2\left(\frac{-25}{4}\right)= -10\)
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