How do you simplify and write #((3x^-2 y^3) /(2xy))^-2# with positive exponents?

2 Answers
Jun 3, 2016

Answer:

#(4x^6)/(9y^4)#

Explanation:

We start with #((3x^-2y^3)/(2xy))^-2#. The first thing I like to do is to simplify everything that I can, which means getting rid of that external #-2#. To do that, we just need to distribute it in. That means we now have #(3x^-2y^3)^-2/(2xy)^-2# or #(1/9x^4y^-6)/(1/4x^-2y^-2)#.

First thing's first, let's get rewrite #(1/9)/(1/4)#. To solve a multiplication problem with fractions, we just need to mutiply the recipricol. For us, that means #1/9*4/1#. Multiply straight across to give us #4/9#.

Now, in order to change the sign of the power, we need to move it into the numerator denominator. What I mean is, if we have a negative power in the denominator, like #1/x^-2#, we need to move the component to to the numerator to change the sign. If we do that, we get #x^2/1#, which is now positive.

For this problem, that means moving the components around, like this: #4/9(x^4*x^2*y^2*y^-6)#. Now, you may notice that I haven't moved #y^-6#. that's because I want to combine like terms before I do that. So, #x^4*x^2# becomes #x^6# because exponents add when the bases are multiplied. That also means that #y^2*y^-6# is #y^4#.

We should now have #(4x^6y^-4)/9#. Let's chnage the sign on that #y^-4# to get #(4x^6)/(9y^4)#. That's it! We are done.

Jun 3, 2016

Answer:

#(4x^6)/(9y^4) #

Explanation:

#((3x^-2y^3)/(2xy))^-2#

There is a nifty rule when working with indices which states that if a fraction is raised to a negative index, if you find the reciprocal of the fraction the index becomes positive:

#(x/y)^-m " becomes " (y/x)^(+m)#

Therefore in the question posed, we can get the following:

#((2xy)/(3x^-2y^3))^2#

Now square everything in the bracket:

#(4x^2y^2)/(9x^-4y^6)#

Simplifying the indices gives the final answer:

#(4x^6)/(9y^4) " note that 2 -(-4) = 6"#