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!
