With any equation you are allowed to take the logarithm of both sides so
5^c=125 now take the logarithm base 5 to both sides giving.
log5(5^c)=log5(125), using the rule log(a^b)=a*log(b) you get
c log5(5)=log5(125), since log5(5)=1
c=log5(125)
1) (3-x) x (3+x)
2) 5(5-x) x (5+x)
3) (3v-wy) x (3v+wy)
4) 2(k-m) x (k+m) x (k^2+m^2)
5) (ab-4c) x (ab+4c)
Brainliest?
