How do you solve #16^ { v - 2} \cdot \frac { 1} { 4} = ( \frac { 1} { 4} ) ^ { - 3v }#?

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Answer 1 · 2017-04-20T21:13:30

#v=-5#

Rewrite #16# and #1/4# as powers of #4#:

#(4^2)^(v-2)*4^-1=(4^-1)^(-3v)#

Using #(a^b)^c=a^(bc)#:

#4^(2v-4)*4^-1=4^(3v)#

Combining the left using #a^b*a^c=a^(b+c)#:

#4^(2v-5)=4^(3v)#

Since the bases are equal, their powers are as well:

#2v-5=3v#

#v=-5#