Question

Simplify:

\[\frac{(16a^6b^{-3}c^5d^9)^{1/4}}{(64a^-4b^-4c^7d^2)^{1/4}}\]

Answer 1

\[\frac{(16a^6b^{-3}c^5d^9)^{1/4}}{(64a{^-4}b^{-4}c^7d^2)^{1/4}}\]

Answer 2

In the numerator it is d^9 and denominator is d^2

I got:
\[\frac{a^2d \sqrt[4]{4a^2bd^3c^2}}{2c}\]

Answer 3

\[\frac{(16a^6b^{-3}c^5d^9)^{1/4}}{(64a^{-4}b^{-4}c^7d^2)^{1/4}}\]\[\large =\frac{(2a^{\frac{6}{4}}b^{\frac{-3}{4}}c^{\frac{5}{4}}d^{\frac{9}{4}})}{(2^{\frac{6}{4}}a^{-1}b^{-1}c^{\frac{7}{4}}d^{\frac{2}{4}})^{1/4}}\]\[\large =2^{(1-\frac{6}{4})}a^{\frac{6}{4}-(-1)}b^{\frac{-3}{4}-(-1)}c^{(\frac{5}{4}-\frac{7}{4})}d^{(\frac{9}{4}-\frac{2}{4})}\]\[\large =2^{-\frac{1}{2}}a^{\frac{5}{2}}b^{\frac{1}{4}}c^{\frac{-1}{2}}d^{\frac{7}{4}}\]\[\large =\frac{a^{\frac{5}{2}}b^{\frac{1}{4}}d^{\frac{7}{4}}}{2^{\frac{1}{2}}c^{\frac{1}{2}}}\]

Answer 4

I am just wondering why you have 2^{1/2}

Answer 5

64 = 2^6
16 = 2^4
(16/64)^(1/4) = 2^{ (4-6) x (1/4)} = 2^(-1/2)

Answer 6

Okay, thank you!