What is the inverse of y = 3log_2 (4x)-2 ?

1 Answer
Nov 10, 2015

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

Explanation:

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!