Question #13eae

Question

Question #13eae

1 Answer

Impact of this question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-19T02:20:03

${(\frac{x^{3} y^{4}}{x^{- 2} y^{- 4}})}^{- 1} = \frac{1}{x^{5} y^{8}} = x^{- 5} y^{- 8}$ $((x^3y^4)/(x^(-2)y^(-4)))^-1=1/(x^5y^8)=x^-5y^-8$

Explanation:

We will use the following properties of exponents:

$x^{a} \cdot x^{b} = x^{a + b}$ $x^a*x^b = x^(a+b)$
${(x^{a})}^{b} = x^{a b}$ $(x^a)^b = x^(ab)$
${(x y)}^{a} = x^{a} y^{a}$ $(xy)^a = x^ay^a$
$x^{0} = 1$ $x^0 = 1$ for all $x \neq 0$ $x!=0$

With those, for $x, y \neq 0$ $x, y != 0$ we have

${(\frac{x^{3} y^{4}}{x^{- 2} y^{- 4}})}^{- 1} = \frac{{(x^{3})}^{- 1} {(y^{4})}^{- 1}}{{(x^{- 2})}^{- 1} {(y^{- 4})}^{- 1}}$ $((x^3y^4)/(x^(-2)y^(-4)))^-1 = ((x^3)^(-1)(y^4)^(-1))/((x^(-2))^(-1)(y^(-4))^(-1))$

$= \frac{x^{3 \cdot - 1} y^{4 \cdot - 1}}{x^{- 2 \cdot - 1} y^{- 4 \cdot - 1}}$ $=(x^(3*-1)y^(4*-1))/(x^(-2*-1)y^(-4*-1))$

$= \frac{x^{- 3} y^{- 4}}{x^{2} y^{4}}$ $=(x^-3y^-4)/(x^2y^4)$

$= \frac{(x^{- 3} y^{- 4}) (x^{3} y^{4})}{(x^{2} y^{4}) (x^{3} y^{4})}$ $=((x^-3y^-4)(x^3y^4))/((x^2y^4)(x^3y^4))$

$= \frac{x^{- 3 + 3} y^{- 4 + 4}}{x^{2 + 3} y^{4 + 4}}$ $=(x^(-3+3)y^(-4+4))/(x^(2+3)y^(4+4))$

$= \frac{x^{0} y^{0}}{x^{5} y^{8}}$ $=(x^0y^0)/(x^5y^8)$

$= \frac{1 \cdot 1}{x^{5} y^{8}}$ $=(1*1)/(x^5y^8)$

$= \frac{1}{x^{5} y^{8}}$ $=1/(x^5y^8)$

Note that we can also multiply this by $\frac{x^{- 5} y^{- 8}}{x^{- 5} y^{- 8}}$ $(x^-5y^-8)/(x^-5y^-8)$ to find it to be equivalent to $x^{- 5} y^{- 8}$ $x^-5y^-8$

Science

Math

Humanities

... and beyond

Question #13eae

1 Answer

Explanation:

Related questions

Impact of this question