Hello, bethany3168!
\(\displaystyle \L \left(\frac{3w^2}{w^4x^3}\right)^{-2}\)
I started this one by breaking it down to: .\(\displaystyle \L \frac{(3w^2)^{-2}}{(w^4x^3)^{-2}} \;= \;\frac{3^{-2}w^{-4}}{w^{-8}x^{-6}}\) . . . . right!
\(\displaystyle 3^{-2}\) is -9 . . . . no
This is where I feel I got stuck.
I have: \(\displaystyle (-9)(w^{-4})(w^{-8})(x^{-6})\) . . . . you lost the division
Go back to:
.\(\displaystyle \L\frac{3^{-2}\cdot w^{-4}}{w^{-8}\cdot x^{-6}}\)
Now, anything with a negative exponent gets "moved".
. . If it is in the numerator, it moves to the denominator.
. . If it is in the denominator, it moves to the numerator.
In both cases, remember to change the exponent to positive.
So we have:
.\(\displaystyle \L \frac{w^8x^6}{3^2w^4}\)
Now you can combine the \(\displaystyle w\)'s and simplify:
. \(\displaystyle \L\frac{w^4x^6}{9}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
By the way, did you know that a negative exponent on a fraction
. . will "flip" the fraction?
Example:
. \(\displaystyle \L\left(\frac{a}{b}\right)^{-3} = \;\left(\frac{b}{a}\right)^3\)
Simplify it the 'normal way':
.\(\displaystyle \L\left(\frac{a}{b}\right)^{-3} = \;\frac{a^{-3}}{b^{-3}} \;= \; \frac{b^3}{a^3}\)
But this is the same as:
. \(\displaystyle \L\left(\frac{a}{b}\right)^{-3} = \;\left(\frac{b}{a}\right)^3 \;= \;\frac{b^3}{a^3}\)
Your problem could hae been:
. \(\displaystyle \L \left(\frac{3w^2}{w^4x^3}\right)^{-2} = \;\left(\frac{w^4x^3}{3w^2}\right)^2\)
. . and eliminate the negative exponents in the very first step!
