+153 Asuming that p and q are nonzero simplify: $\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}$
+9481 We could do it pretty quickly by applying the exponent rules....but if that's confusing or if you're not super comfortable using exponent rules, we can go through the first part like this:
First let's just look at (pq3)3 :
\({(pq^3)^3}\ =\ (pq^3)(pq^3)(pq^3)\)
Imagine replacing every q3 with qqq. How many p's would there be? 3. How many q's would there be? 9. So...
\( {(pq^3)^3}\ =\ p^3q^9\)
Next let's look at (4p2q)2 :
\((4p^2q)^2\ =\ (4p^2q)(4p^2q)\)
In this case, how many p's are there? 4. How many q's are there? 2. How many 4's are there? 2. So...
\({(4p^2q)^2}\ =\ 16p^4q^2\)
Now let's look at (2pq2)3 :
\((2pq^2)^3\ =\ (2pq^2) (2pq^2) (2pq^2)\)
Again let's ask how many p's, q's , and 2's there are to get that...
\( {(2pq^2)^3}\ =\ 8p^3q^6\)
And so...
\(\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}\ =\ \dfrac{(p^3q^9)(16p^4q^2)}{8p^3q^6}\)
There are 3 p's in the first set of parenthesees in the numerator and 4 p's in the second. That makes 7 p's total in the numerator. There are 9 q's in the first set of parenthesees in the numerator and 2 q's in the second. That makes 11 q's total in the numerator. I'm not going to write explanations for these next steps... so feel free to ask if anything confuses you!!
\(\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}\ =\ \dfrac{(p^3q^9)(16p^4q^2)}{8p^3q^6}\\~\\ \phantom{\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}}\ =\ \dfrac{16p^7q^{11}}{8p^3q^6} \\~\\ \phantom{\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}}\ =\ \dfrac{16}{8}\cdot\dfrac{p^7}{p^3}\cdot\dfrac{q^{11}}{q^6}\\~\\ \phantom{\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}}\ =\ 2\cdot p^4\cdot q^5\\~\\ \phantom{\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}}\ =\ 2p^4q^5\)
Anytime you don't know how to handle an exponent, it can help to rewrite it as repeated multiplication....I used to always do this until I became super familiar with exponent rules...and I still have to do it sometimes!!
Hope this helps!
