First, switch #y# and #x# in your equation:

#x = 3 log_2(4y) - 2#

Now, solve this equation for #y#:

#x = 3 log_2 (4y) - 2#

#<=> x + 2 = 3 log_2(4y)#

#<=> (x+2)/3 = log_2(4y)#

The inverse function of #log_2(a)# is #2^a#, so apply this operation to both sides of the equation to get rid of the logarithm:

#<=> 2^((x+2)/3) = 2^(log_2(4y))#

#<=> 2^((x+2)/3) = 4y#

Let's simplify the expression on the left side using the power rules #a^n * a^m = a^(n+m)# and #a^(n*m) = (a^n)^m#:

#2^((x+2)/3) = 2^(x/3 + 2/3) = 2^(x/3) * 2^(2/3) = 2^(x/3) * (2^2)^(1/3) = 4^(1/3) * 2^(x/3)#

Let's get back to our equation:

#2^((x+2)/3) = 4y#

#<=> 4^(1/3) * 2^(x/3) = 4y#

#<=> 4^(1/3)/4 * 2^(x/3) = y#

#<=> 4^(-2/3) * 2^(x/3) = y#

You are done. The only thing left to do is replacing #y# with #f^(-1)(x)# for a more formal notation:

for

#f(x) = 3 log_2(4x) - 2#,

the inverse function is

#f^(-1)(x) = 4^(-2/3) * 2^(x/3)#.

Hope that this helped!