Solve the following a) \( 4 \cdot 3^{x}+3^{x}=15 \) b) \( 2^{x+1}+2^{x-1}=5 \) c) \( 3\left(3^{x+2}+3^{x-1}\right)=84 \) d) \( 4^{x}+4 \cdot 2^{x}-5=0 \) e) \( 6.10^{x}+10^{x+2}-106=0 \)

Let's solve each equation step by step. --- ### **a) \( 4 \cdot 3^{x} + 3^{x} = 15 \)** **Solution:** 1. Combine like terms: \[ 4 \cdot 3^{x} + 3^{x} = 5 \cdot 3^{x} \] 2. Set the equation equal to 15: \[ 5 \cdot 3^{x} = 15 \] 3. Divide both sides by 5: \[ 3^{x} = 3 \] 4. Solve for \( x \): \[ x = 1 \] **Answer:** \( x = 1 \) --- ### **b) \( 2^{x+1} + 2^{x-1} = 5 \)** **Solution:** 1. Rewrite the exponents: \[ 2^{x+1} = 2 \cdot 2^{x} \quad \text{and} \quad 2^{x-1} = \frac{1}{2} \cdot 2^{x} \] 2. Substitute back into the equation: \[ 2 \cdot 2^{x} + \frac{1}{2} \cdot 2^{x} = 5 \] 3. Factor out \( 2^{x} \): \[ \left(2 + \frac{1}{2}\right) \cdot 2^{x} = 5 \quad \Rightarrow \quad \frac{5}{2} \cdot 2^{x} = 5 \] 4. Solve for \( 2^{x} \): \[ 2^{x} = 2 \] 5. Therefore: \[ x = 1 \] **Answer:** \( x = 1 \) --- ### **c) \( 3\left(3^{x+2} + 3^{x-1}\right) = 84 \)** **Solution:** 1. Divide both sides by 3: \[ 3^{x+2} + 3^{x-1} = 28 \] 2. Express in terms of \( 3^{x-1} \): \[ 3^{x+2} = 3^{x-1} \cdot 3^{3} = 27 \cdot 3^{x-1} \] \[ 27 \cdot 3^{x-1} + 3^{x-1} = 28 \] 3. Factor out \( 3^{x-1} \): \[ 28 \cdot 3^{x-1} = 28 \] 4. Divide by 28: \[ 3^{x-1} = 1 \quad \Rightarrow \quad x-1 = 0 \quad \Rightarrow \quad x = 1 \] **Answer:** \( x = 1 \) --- ### **d) \( 4^{x} + 4 \cdot 2^{x} - 5 = 0 \)** **Solution:** 1. Rewrite \( 4^{x} \) as \( (2^2)^x = 2^{2x} \). 2. Let \( y = 2^{x} \). Then \( 2^{2x} = y^2 \). 3. Substitute into the equation: \[ y^2 + 4y - 5 = 0 \] 4. Solve the quadratic equation: \[ y = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2} \] \[ y = 1 \quad \text{or} \quad y = -5 \] 5. Since \( y = 2^{x} \) must be positive: \[ 2^{x} = 1 \quad \Rightarrow \quad x = 0 \] **Answer:** \( x = 0 \) --- ### **e) \( 6 \cdot 10^{x} + 10^{x+2} - 106 = 0 \)** **Solution:** 1. Rewrite \( 10^{x+2} \) as \( 10^2 \cdot 10^{x} = 100 \cdot 10^{x} \). 2. Substitute back into the equation: \[ 6 \cdot 10^{x} + 100 \cdot 10^{x} - 106 = 0 \] 3. Combine like terms: \[ 106 \cdot 10^{x} - 106 = 0 \] 4. Solve for \( 10^{x} \): \[ 106 \cdot 10^{x} = 106 \quad \Rightarrow \quad 10^{x} = 1 \] 5. Therefore: \[ x = 0 \] **Answer:** \( x = 0 \) --- ### **Summary of Solutions:** - **a)** \( x = 1 \) - **b)** \( x = 1 \) - **c)** \( x = 1 \) - **d)** \( x = 0 \) - **e)** \( x = 0 \)