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The given equation is:

\[
2^{2x} - 20 \cdot 2^x + 64 = 0
\]

Let's solve this step by step.

### Step 1: Substitution
Let \( y = 2^x \). This transforms the equation into a quadratic form:
\[
y^2 - 20y + 64 = 0
\]

### Step 2: Solve the quadratic equation
We now solve the quadratic equation:
\[
y^2 - 20y + 64 = 0
\]

Using the quadratic formula:
\[
y = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(64)}}{2(1)}
\]
\[
y = \frac{20 \pm \sqrt{400 - 256}}{2}
\]
\[
y = \frac{20 \pm \sqrt{144}}{2}
\]
\[
y = \frac{20 \pm 12}{2}
\]

This gives two solutions:
\[
y_1 = \frac{20 + 12}{2} = 16
\]
\[
y_2 = \frac{20 - 12}{2} = 4
\]

### Step 3: Back-substitute \( y = 2^x \)
Now, we back-substitute \( y = 2^x \).

For \( y_1 = 16 \):
\[
2^x = 16
\]
\[
x = 4 \quad (\text{since } 2^4 = 16)
\]

For \( y_2 = 4 \):
\[
2^x = 4
\]
\[
x = 2 \quad (\text{since } 2^2 = 4)
\]

### Solution
The solution set is:
\[
x = 2 \quad \text{and} \quad x = 4
\]

The solution set (himpunan penyelesaian) is:
\[
\{2, 4\}
\]

Would you like further details or have any questions?

Here are some related questions to expand on the solution:
1. How is the quadratic formula derived and when can it be used?
2. What are other methods to solve quadratic equations?
3. Can all exponential equations be transformed into quadratic equations?
4. How do you solve higher-degree polynomial equations?
5. What is the significance of substitution in simplifying equations?

**Tip**: Always check if your equation can be simplified using substitutions or factoring—it often makes solving much easier!

Himpunan penyelesaian dari 2^{2x} - 20 · 2^x + 64 = 0 adalah …

This page provides a step-by-step solution to the exponential equation 2^(2x) - 20 · 2^x + 64 = 0. Learn how to use substitution to convert it into a quadratic equation, followed by solving the quadratic using the quadratic formula. Suitable for students in Grades 10-12.

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