What is the value of #b# that would make this equation true #b\root[ 3] { 64a ^ { \frac { b } { 2} } } = ( 4\sqrt { 3} a ) ^ { 2}#?

1 Answer
Jan 11, 2018

#b=12#

Explanation:

There are several ways to see this. Here's one:

Given:

#b root(3)(64a^(b/2)) = (4sqrt(3)a)^2#

Cube both sides to get:

#64 b^3 a^(b/2) = (4sqrt(3)a)^6 = 4^6 * 3^3 a^6#

Equating powers of #a# we have:

#b/2 = 6#

Hence:

#b = 12#

To check, divide both ends by #4^3 = 64# to get:

#b^3 a^(b/2) = 4^3 * 3^3 a^6 = 12^3 a^6#

So looking at the coefficient of #a^(b/2) = a^6#, we have #b^3 = 12^3# and hence #b=12# works.