1. **Problem statement:** Find the inverse function value $f^{-1}(2)$ for the function $$f(x) = \frac{2^x - (\frac{1}{2})^x}{2}.$$\n\n2. **Recall the function and inverse definition:** The function is given by $$f(x) = \frac{2^x - 2^{-x}}{2}.$$ We want to find $x$ such that $f(x) = 2$. This means solving $$\frac{2^x - 2^{-x}}{2} = 2.$$\n\n3. **Multiply both sides by 2:** $$2^x - 2^{-x} = 4.$$\n\n4. **Substitute $y = 2^x$:** Then $2^{-x} = \frac{1}{y}$. The equation becomes $$y - \frac{1}{y} = 4.$$\n\n5. **Multiply both sides by $y$ to clear the denominator:** $$y^2 - 1 = 4y.$$\n\n6. **Rearrange to form a quadratic equation:** $$y^2 - 4y - 1 = 0.$$\n\n7. **Solve the quadratic equation using the quadratic formula:** $$y = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}.$$\n\n8. **Since $y = 2^x > 0$, both roots are positive, but we check which is valid:**\n- $2 + \sqrt{5} > 0$\n- $2 - \sqrt{5} \approx 2 - 2.236 < 0$ (not valid since $y>0$)\n\nSo, $$y = 2 + \sqrt{5}.$$\n\n9. **Find $x$ by taking logarithm base 2:** $$x = \log_2\left(2 + \sqrt{5}\right).$$\n\n**Final answer:** $$f^{-1}(2) = \log_2\left(2 + \sqrt{5}\right).$$