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

The answer I came up with was: $\displaystyle 54x^{13}y$

but I feel this answer is incorrect. Please, what is the correct answer and how does one get it?

Thanks in advance!

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- Mar 16th 2008, 02:15 PMmt_lapinSimplify the expression
$\displaystyle (18x^3y^4)^{-1}(3x^4y)^4$

The answer I came up with was: $\displaystyle 54x^{13}y$

but I feel this answer is incorrect. Please, what is the correct answer and how does one get it?

Thanks in advance! - Mar 16th 2008, 02:24 PMMoo
Hello,

There are three main things to do :

- $\displaystyle (a^b)^c = a^{bc}$

- $\displaystyle a^b a^c = a^{b+c}$

- $\displaystyle (ab)^c = a^c b^c$

And note that 18 is 3²*2 - Mar 16th 2008, 02:30 PMtopher0805
$\displaystyle

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

$

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

So you would have:

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

Factor out a 9:

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

Factor out an $\displaystyle x^3$:

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

Factor out a $\displaystyle y^4$:

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