Question

Simplify the expression and eliminate any negative exponent(s). Assume that y and z denote positive numbers.
(16y^-2z^4)^1/2 over (8y^3z^-6)^-1/3

Answer 1

\[\frac{(16y^{-2}z^4)^\frac{1}{2}}{(8y^3z^{-6})^{\frac{1}{3}}}\]

Answer 2

ok lets get rid of the fractional exponents first by multiplying every exponent in the numerator by 1/2 and every one in the denominator by 1/3
also know that
\[16^{\frac{1}{2}}=\sqrt{16}=4\]\]

Answer 3

and likewise
\[8^{\frac{1}{3}}=\sqrt[3]{8}=2\]

Answer 4

so the numerator will be
\[4y^{-1}z^2\] and the denominator will be \[2yz^{-3}\]

Answer 5

our fraction is now
\[\frac{4y^{-1}z^2}{2yz^{-2}}\]

Answer 6

ok whats next

Answer 7

to make the negative exponents positive, if it is up bring it down and if it is down bring it up so you get
\[\frac{4z^2z^2}{2yy}\]which gives
\[\frac{2z^4}{y^2}\]

Answer 8

i also of course divided 4 by 2 to get 2 in the numerator

Answer 9

gotcha. whats the final

Answer 10

notice that the
\[z^{-2}\] in the denominator came up to the numerator as
\[z^2\] and the
\[y^{-1}\] in the numerator came to the denominator as
\[y\]

Answer 11

"final answer" is
\[\frac{2z^4}{y^2}\]

Answer 12

\[\color{green}{\text{hello karnak}}\]