Suppose the functions \( f, g, h \) and \( l \) are defined as follows: \[ \begin{aligned} f(x) & =4 x^{2}-5 x+1 \\ g(x) & =2 \sqrt{2-\frac{x}{2}}-x \\ h(x) & =-\frac{1}{2} x+3 \\ l(x) & =\log _{4}(x+3)-\log _{4}(x-2)\end{aligned} \] \[ \begin{array}{l}\text { Solve the equation } 4^{h(x)}=8\end{array} \]

To solve the equation \( 4^{h(x)} = 8 \), let's follow these steps: 1. **Understand the Function \( h(x) \):** \[ h(x) = -\frac{1}{2}x + 3 \] 2. **Set Up the Equation:** \[ 4^{h(x)} = 8 \] 3. **Express Both Sides with the Same Base:** Notice that both 4 and 8 can be expressed as powers of 2: \[ 4 = 2^2 \] \[ 8 = 2^3 \] So, rewrite the equation: \[ (2^2)^{h(x)} = 2^3 \] \[ 2^{2h(x)} = 2^3 \] 4. **Equate the Exponents:** Since the bases are the same, set the exponents equal to each other: \[ 2h(x) = 3 \] 5. **Solve for \( h(x) \):** \[ h(x) = \frac{3}{2} \] 6. **Substitute \( h(x) \) Back into Its Definition:** \[ -\frac{1}{2}x + 3 = \frac{3}{2} \] 7. **Solve for \( x \):** \[ \begin{align*} -\frac{1}{2}x + 3 &= \frac{3}{2} \\ -\frac{1}{2}x &= \frac{3}{2} - 3 \\ -\frac{1}{2}x &= -\frac{3}{2} \\ x &= 3 \end{align*} \] 8. **Check the Solution:** Ensure that \( x = 3 \) is within the domain of \( h(x) \) (which it is, since \( h(x) \) is defined for all real numbers). **Final Answer:** \[ x = 3 \]