Answer:
4
2
3
1
Step-by-step explanation:
- [tex]$ (a^x)^{\frac{1}{y}} = a^\frac{x}{y} $[/tex]
- [tex]$ \frac{a^x}{a^y} = a^{x - y} $[/tex]
1) [tex](p^6.q^{{\frac{3}{2}}})^{{\frac{1}{3}}}[/tex]
Therefore, [tex]$ p^{\frac{6}{3}}.q^{\frac{3}{2}.\frac{1}{3}} $[/tex]
[tex]$ \implies p^2.q^{\frac{1}{2}} $[/tex]
2) [tex]$ \frac{p^5}{p^{-3}q^{-4}} $[/tex]
[tex]$ = p^5.p^3.q^4 = p^8.q^4 $[/tex]
Now, [tex]$ (p^8q^4)^{\frac{1}{4}} $[/tex]
[tex]$ \implies p^{\frac{8}{4}}.q^{\frac{4}{4}} $[/tex]
[tex]$ = p^2. q $[/tex]
3) [tex]$ (\frac{p^2.q^7}{q^4} )^{\frac{1}{2} $[/tex]
[tex]$ \implies p^{{\frac{2}{2}}}.q^{\frac{3}{2}} $[/tex]
[tex]$ \implies pq^{\frac{3}{2}} $[/tex]
4) [tex]$ \frac{(pq^3)^{{\frac{1}{2}}}}{(pq)^{{\frac{-1}{2}}}} $[/tex]
[tex]$ \implies p^{{\frac{1}{2}}}q^{{\frac{3}{2}}}. p^{{\frac{1}{2}}}.q^{{\frac{1}{2}}}} $[/tex]
[tex]$ = pq^2 $[/tex]
Hence, the answers.
