Simplify the expression

**mt_lapin** · March 16th 2008, 02:15 PM

$(18x^3y^4)^{-1}(3x^4y)^4$

The answer I came up with was: $54x^{13}y$
but I feel this answer is incorrect. Please, what is the correct answer and how does one get it?

Thanks in advance!

**Moo** · March 16th 2008, 02:24 PM

Hello,

There are three main things to do :

- $(a^b)^c = a^{bc}$
- $a^b a^c = a^{b+c}$
- $(ab)^c = a^c b^c$

And note that 18 is 3²*2

**topher0805** · March 16th 2008, 02:30 PM

$<br /> (18x^3y^4)^{-1}(3x^4y)^4<br />$

Recall that $(18x^3y^4)^{-1} = \frac {1}{(18x^3y^4)}$

So you would have:

$\frac {81x^{16}y^4}{18x^3y^4}$

Factor out a 9:

$\frac {9x^{16}y^4}{2x^3y^4}$

Factor out an $x^3$ :

$\frac {9x^{13}y^4}{2y^4}$

Factor out a $y^4$ :

$\frac {9x^{13}}{2}$

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