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

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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:**
$$ x = 3 $$
The solution to the equation $$ 4^{h(x)} = 8 $$ is $$ x = 3 $$.

Suppose the functions and are defined as follows:

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