1. **State the problem:** Solve the equation $ (2^x)^2 - 3 \times 2^x + 2 = 0 $. 2. **Substitution:** Let $ y = 2^x $. Then the equation becomes: $$ y^2 - 3y + 2 = 0 $$ 3. **Solve the quadratic equation:** Factorize: $$ y^2 - 3y + 2 = (y - 1)(y - 2) = 0 $$ So, $$ y = 1 \quad \text{or} \quad y = 2 $$ 4. **Back-substitute for $x$:** Recall $ y = 2^x $, so: - If $ y = 1 $, then $ 2^x = 1 $ which implies $ x = 0 $ because $ 2^0 = 1 $. - If $ y = 2 $, then $ 2^x = 2 $ which implies $ x = 1 $ because $ 2^1 = 2 $. 5. **Final answer:** $$ x = 0 \quad \text{or} \quad x = 1 $$