How do you simplify $\frac{6^{- 2} x^{3} y^{- 4}}{2^{- 4} x^{3} y^{- 4}}$ $(6^-2x^3y^-4)/(2^-4x^3y^-4)$ ?

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Answer 1 · 2018-05-25T21:41:17

$\frac{4}{9}$ $4/9$

The $x$ $x$ and $y$ $y$ terms have the same exponent on the top and bottom, so they will essentially cancel each other out. We're left with

$\frac{6^{- 2}}{2^{- 4}}$ $6^-2/2^-4$

We can rewrite this as being equal to

$\frac{\frac{1}{6^{2}}}{\frac{1}{2^{4}}}$ $(1/6^2)/(1/2^4)$

Which can be further simplified as follows

$\Rightarrow \frac{\frac{1}{36}}{\frac{1}{16}}$ $=>(1/36)/(1/16)$

$\Rightarrow \frac{1}{36} \cdot 16$ $=>1/36*16$

$\Rightarrow \frac{16}{36}$ $=>16/36$

$\Rightarrow \frac{4}{9}$ $=>4/9$

Hope this helps!

How do you simplify (6^-2x^3y^-4)/(2^-4x^3y^-4)6−2x3y−42−4x3y−4(6^-2x^3y^-4)/(2^-4x^3y^-4)?