How do you solve #16^ { v - 2} \cdot \frac { 1} { 4} = ( \frac { 1} { 4} ) ^ { - 3v }#?
1 Answer
Apr 20, 2017
Explanation:
Rewrite
#(4^2)^(v-2)*4^-1=(4^-1)^(-3v)#
Using
#4^(2v-4)*4^-1=4^(3v)#
Combining the left using
#4^(2v-5)=4^(3v)#
Since the bases are equal, their powers are as well:
#2v-5=3v#
#v=-5#