Answer:
Option B is correct.
[tex]\frac{81m^2n^5}{8}[/tex] is equivalent to [tex]\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}[/tex]
Step-by-step explanation:
Given expression: [tex]\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}[/tex]
Using exponents power:
- [tex](ab)^n = a^nb^n[/tex]
- [tex](a^n)^m = a^{nm}[/tex]
- [tex]a^m \cdot a^n = a^{m+n}[/tex]
Given: [tex]\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}[/tex]
Apply exponent power :
⇒ [tex]\frac{3^4 (m^{-1})^4(n^2)^4}{2^3(m^{-2})^3 n^3}[/tex]
⇒ [tex]\frac{81 m^{-4}n^8}{8m^{-6}n^3} = \frac{81 m^{-4} \cdot m^6 n^8 \cdot n^{-3}}{8}[/tex]
⇒[tex]\frac{81 m^{-4+6} n^{8-3}}{8} = \frac{81 m^2 n^5}{8} = \frac{81m^2 n^5}{8}[/tex]
Therefore, the expression which is equivalent to [tex]\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}[/tex] is, [tex]\frac{81m^2 n^5}{8}[/tex]
