Question

(m^3n)^-3/m^-7n^8

Answer 1

(m^3n)^-3/m^-7n^8

Answer 2

omg this is confusing i gotta rewrite it
\( \huge{ (m^{3n})^{-3} \over (m^{-7n})^{8}} \)

Answer 3

is that it?

Answer 4

lol.. yes

Answer 5

ok so the first thing we would do is apply the law of exponents which is
\(\large(a^n)^m=a^{m*n} \)

Answer 6

so in the numerator we have
\( \large(m^{3n})^{-3}=m^{3n*-3}=? \)

Answer 7

where would the -3 go?

Answer 8

like i showed u u keep the base m and mutiply the exponents 3n*(-3)
so \(\large(m^{3n})^{-3}=m^{3n*(-3)}=m^{-9n} \)

Answer 9

and lets go to the denominator
\(\large (m^{-7n})^{8}=m^{-7n*8}=m^{-56n} \)

Answer 10

so lets put the pieces together, so we have
\(\huge{m^{-9n} \over m^{-56n}} \)

Answer 11

when dividing with exponents. If the bases are the same then we keep the base and subtract the exponents

Answer 12

my bad.. so its gonna be m^56n/m^9n

Answer 13

\( \huge m^{-9n-(-56n)} =? \)

Answer 14

\( \large m^{56n-9n}=? \)

Answer 15

1/m^47n

Answer 16

hmmmmm
\( m^{47n}\) is the answer

Answer 17

ok. thanks!