The function $g$ is defined as $g(x) = \frac{2x-3}{x}$. We need to find $g^{-1}(\frac{1}{2})$.

1. Problem Description

The function $g$ is defined as $g(x) = \frac{2x-3}{x}$. We need to find $g^{-1}(\frac{1}{2})$.

2. Solution Steps

First, let $y = g(x)$. Then $y = \frac{2x-3}{x}$.

To find the inverse function $g^{-1}(y)$, we need to solve for $x$ in terms of $y$.

$y = \frac{2x-3}{x}$

$xy = 2x - 3$

$xy - 2x = -3$

$x(y-2) = -3$

$x = \frac{-3}{y-2}$

$x = \frac{3}{2-y}$

Thus, $g^{-1}(y) = \frac{3}{2-y}$.

Now, we can find $g^{-1}(\frac{1}{2})$ by substituting $y = \frac{1}{2}$ into the expression for $g^{-1}(y)$:

$g^{-1}(\frac{1}{2}) = \frac{3}{2-\frac{1}{2}}$

$g^{-1}(\frac{1}{2}) = \frac{3}{\frac{4}{2}-\frac{1}{2}}$

$g^{-1}(\frac{1}{2}) = \frac{3}{\frac{3}{2}}$

$g^{-1}(\frac{1}{2}) = 3 \cdot \frac{2}{3}$

$g^{-1}(\frac{1}{2}) = 2$

3. Final Answer

$g^{-1}(\frac{1}{2}) = 2$