\( \begin{array}{ll}\text { c) } 2^{2 x}-2.2^{x}+1=0 & \text { d) } 2^{2 x+2}-5.2^{x}+1=0 \\ \text { e) } 2^{2 x+1}-2^{x}=0 & \text { f) } 2^{2 x}=2^{2}\left(2^{x}+8\right) \\ \text { g) } 9^{x+1}-10.3^{x}+1=0 & \text { h) } 4^{x}+2^{x}=2^{3+x}+2^{3}\end{array} \)

Sure, let's solve each of the given exponential equations step by step. --- ### **Problem c)** \[ 2^{2x} - 2 \cdot 2^{x} + 1 = 0 \] **Solution:** 1. **Substitution:** Let \( y = 2^x \). Then \( 2^{2x} = y^2 \). 2. **Rewrite the equation:** \[ y^2 - 2y + 1 = 0 \] 3. **Factor the quadratic:** \[ (y - 1)^2 = 0 \] 4. **Solve for \( y \):** \[ y = 1 \] 5. **Back-substitute \( y = 2^x \):** \[ 2^x = 1 \] 6. **Solve for \( x \):** \[ x = 0 \] **Answer:** \( x = 0 \) --- ### **Problem d)** \[ 2^{2x+2} - 5 \cdot 2^{x} + 1 = 0 \] **Solution:** 1. **Simplify \( 2^{2x+2} \):** \[ 2^{2x+2} = 4 \cdot 2^{2x} \] 2. **Substitution:** Let \( y = 2^x \). Then \( 2^{2x} = y^2 \). 3. **Rewrite the equation:** \[ 4y^2 - 5y + 1 = 0 \] 4. **Solve the quadratic equation:** \[ y = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8} \] - \( y = \frac{8}{8} = 1 \) - \( y = \frac{2}{8} = \frac{1}{4} \) 5. **Back-substitute \( y = 2^x \):** - For \( y = 1 \): \( 2^x = 1 \) ⇒ \( x = 0 \) - For \( y = \frac{1}{4} \): \( 2^x = 2^{-2} \) ⇒ \( x = -2 \) **Answer:** \( x = 0 \) and \( x = -2 \) --- ### **Problem e)** \[ 2^{2x+1} - 2^{x} = 0 \] **Solution:** 1. **Factor out \( 2^x \):** \[ 2^x (2^{x+1} - 1) = 0 \] - \( 2^x \) is never zero, so set the second factor to zero: 2. **Set \( 2^{x+1} - 1 = 0 \):** \[ 2^{x+1} = 1 \] 3. **Solve for \( x \):** \[ x + 1 = 0 \] ⇒ \( x = -1 \) **Answer:** \( x = -1 \) --- ### **Problem f)** \[ 2^{2x} = 2^2 (2^{x} + 8) \] **Solution:** 1. **Simplify the right-hand side:** \[ 2^2 (2^x + 8) = 4 \cdot 2^x + 32 \] 2. **Substitution:** Let \( y = 2^x \). Then \( 2^{2x} = y^2 \). 3. **Rewrite the equation:** \[ y^2 = 4y + 32 \] 4. **Rearrange to form a quadratic:** \[ y^2 - 4y - 32 = 0 \] 5. **Solve the quadratic equation:** \[ y = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2} \] - \( y = \frac{16}{2} = 8 \) (valid since \( y > 0 \)) - \( y = \frac{-8}{2} = -4 \) (invalid) 6. **Back-substitute \( y = 2^x \):** \[ 2^x = 8 \] ⇒ \( x = 3 \) **Answer:** \( x = 3 \) --- ### **Problem g)** \[ 9^{x+1} - 10 \cdot 3^{x} + 1 = 0 \] **Solution:** 1. **Express \( 9^{x+1} \) in terms of base 3:** \[ 9^{x+1} = 3^{2(x+1)} = 3^{2x + 2} = 9 \cdot 3^{2x} \] 2. **Substitution:** Let \( y = 3^x \). Then \( 3^{2x} = y^2 \). 3. **Rewrite the equation:** \[ 9y^2 - 10y + 1 = 0 \] 4. **Solve the quadratic equation:** \[ y = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm 8}{18} \] - \( y = \frac{18}{18} = 1 \) - \( y = \frac{2}{18} = \frac{1}{9} \) 5. **Back-substitute \( y = 3^x \):** - For \( y = 1 \): \( 3^x = 1 \) ⇒ \( x = 0 \) - For \( y = \frac{1}{9} \): \( 3^x = 3^{-2} \) ⇒ \( x = -2 \) **Answer:** \( x = 0 \) and \( x = -2 \) --- ### **Problem h)** \[ 4^{x} + 2^{x} = 2^{3+x} + 2^{3} \] **Solution:** 1. **Express all terms in base 2:** \[ 4^x = (2^2)^x = 2^{2x} \] \[ 2^{3+x} = 2^x \cdot 2^3 = 8 \cdot 2^x \] \[ 2^3 = 8 \] 2. **Rewrite the equation:** \[ 2^{2x} + 2^x = 8 \cdot 2^x + 8 \] 3. **Substitution:** Let \( y = 2^x \). Then \( 2^{2x} = y^2 \). 4. **Rewrite the equation:** \[ y^2 + y = 8y + 8 \] 5. **Rearrange to form a quadratic:** \[ y^2 - 7y - 8 = 0 \] 6. **Solve the quadratic equation:** \[ y = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2} \] - \( y = \frac{16}{2} = 8 \) (valid) - \( y = \frac{-2}{2} = -1 \) (invalid) 7. **Back-substitute \( y = 2^x \):** \[ 2^x = 8 \] ⇒ \( x = 3 \) **Answer:** \( x = 3 \) --- **Summary of Solutions:** - **c)** \( x = 0 \) - **d)** \( x = 0 \) and \( x = -2 \) - **e)** \( x = -1 \) - **f)** \( x = 3 \) - **g)** \( x = 0 \) and \( x = -2 \) - **h)** \( x = 3 \)