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

Explanation

= \(\frac{(a^2)^{3/4}\times(b^{-3})^{3/4}\times c^{3/4}}{a^{-1} b^4 c^5}\) \(= a^p\times b^q\times c^r\)

= \(\frac{a^{3/2}\times b^{-9/4} \times c^{3/4}}{a^{-1}b^4c^5}\) = \(a^p b^q c^r\) =  \(a^{{3/4}+1} b^{{-9/4}-4} c^{{3/5}-5}\) = \(a^pb^q c^r\)

= \(a^{5/2} b^{-25/4} c^{-17/4}\) = \(a^pb^qc^r\)

therefore, p=5/2, q=-25/4 and r = -17/4

2a + q = \(\frac{5}{2} - 2\times\frac{-25}{4}\)

= -20/2 = -10