(2) \( 3 \cdot 5^{x}+5^{x+1}=40 \) (4) \( \sqrt{9^{x+2}}-3^{x-1}=\frac{26}{27} \) (6) \( 4^{x}+16=10 \cdot 2^{x} \) (8) \( 16^{x}-3 \cdot 4^{x+1}=-32 \) (10) \( 5^{x}-5^{2-x}=24 \) (12) \( 3^{4 x}-12 \cdot 3^{2 x}+27=0 \)

Sure, let's solve each of the given equations step by step. --- ### Problem (2) **Equation:** \[ 3 \cdot 5^{x} + 5^{x+1} = 40 \] **Solution:** 1. Rewrite \(5^{x+1}\) as \(5 \cdot 5^x\): \[ 3 \cdot 5^{x} + 5 \cdot 5^{x} = 40 \] 2. Factor out \(5^x\): \[ 5^{x} (3 + 5) = 40 \] \[ 5^{x} \cdot 8 = 40 \] 3. Solve for \(5^x\): \[ 5^{x} = \frac{40}{8} = 5 \] 4. Since \(5^x = 5^1\), we have: \[ x = 1 \] **Answer:** \[ x = 1 \] --- ### Problem (4) **Equation:** \[ \sqrt{9^{x+2}} - 3^{x-1} = \frac{26}{27} \] **Solution:** 1. Express \(9^{x+2}\) in terms of base 3: \[ 9^{x+2} = (3^2)^{x+2} = 3^{2x + 4} \] 2. Take the square root: \[ \sqrt{3^{2x + 4}} = 3^{x + 2} \] 3. Rewrite the equation: \[ 3^{x + 2} - 3^{x - 1} = \frac{26}{27} \] 4. Factor out \(3^{x - 1}\): \[ 3^{x-1} (3^3 - 1) = \frac{26}{27} \] \[ 3^{x-1} \cdot 26 = \frac{26}{27} \] 5. Solve for \(3^{x-1}\): \[ 3^{x-1} = \frac{1}{27} = 3^{-3} \] 6. Equate the exponents: \[ x - 1 = -3 \] \[ x = -2 \] **Answer:** \[ x = -2 \] --- ### Problem (6) **Equation:** \[ 4^{x} + 16 = 10 \cdot 2^{x} \] **Solution:** 1. Express \(4^x\) as \(2^{2x}\): \[ 2^{2x} + 16 = 10 \cdot 2^{x} \] 2. Let \(y = 2^x\), then \(2^{2x} = y^2\): \[ y^2 + 16 = 10y \] 3. Rearrange the equation: \[ y^2 - 10y + 16 = 0 \] 4. Solve the quadratic equation: \[ y = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2} \] \[ y = 8 \quad \text{or} \quad y = 2 \] 5. Convert back to \(x\): - For \(y = 8\): \[ 2^x = 8 \Rightarrow x = 3 \] - For \(y = 2\): \[ 2^x = 2 \Rightarrow x = 1 \] **Answer:** \[ x = 1 \quad \text{and} \quad x = 3 \] --- ### Problem (8) **Equation:** \[ 16^{x} - 3 \cdot 4^{x+1} = -32 \] **Solution:** 1. Express \(16^x\) and \(4^{x+1}\) in terms of base 4: \[ 16^{x} = (4^2)^x = 4^{2x} \] \[ 4^{x+1} = 4^x \cdot 4 = 4 \cdot 4^x \] 2. Let \(y = 4^x\), then: \[ y^2 - 3 \cdot 4y = -32 \] 3. Rearrange the equation: \[ y^2 - 12y + 32 = 0 \] 4. Solve the quadratic equation: \[ y = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm 4}{2} \] \[ y = 8 \quad \text{or} \quad y = 4 \] 5. Convert back to \(x\): - For \(y = 8\): \[ 4^x = 8 \Rightarrow (2^2)^x = 2^3 \Rightarrow 2^{2x} = 2^3 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2} \] - For \(y = 4\): \[ 4^x = 4 \Rightarrow x = 1 \] **Answer:** \[ x = 1 \quad \text{and} \quad x = \frac{3}{2} \] --- ### Problem (10) **Equation:** \[ 5^{x} - 5^{2 - x} = 24 \] **Solution:** 1. Let \(y = 5^x\), then \(5^{2 - x} = \frac{25}{y}\): \[ y - \frac{25}{y} = 24 \] 2. Multiply both sides by \(y\): \[ y^2 - 25 = 24y \] 3. Rearrange the equation: \[ y^2 - 24y - 25 = 0 \] 4. Solve the quadratic equation: \[ y = \frac{24 \pm \sqrt{576 + 100}}{2} = \frac{24 \pm \sqrt{676}}{2} = \frac{24 \pm 26}{2} \] \[ y = 25 \quad \text{or} \quad y = -1 \] 5. Since \(y = 5^x > 0\), discard \(y = -1\): \[ 5^x = 25 \Rightarrow x = 2 \] **Answer:** \[ x = 2 \] --- ### Problem (12) **Equation:** \[ 3^{4x} - 12 \cdot 3^{2x} + 27 = 0 \] **Solution:** 1. Let \(y = 3^{2x}\), then \(3^{4x} = y^2\): \[ y^2 - 12y + 27 = 0 \] 2. Solve the quadratic equation: \[ y = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2} \] \[ y = 9 \quad \text{or} \quad y = 3 \] 3. Convert back to \(x\): - For \(y = 9\): \[ 3^{2x} = 9 \Rightarrow 3^{2x} = 3^2 \Rightarrow 2x = 2 \Rightarrow x = 1 \] - For \(y = 3\): \[ 3^{2x} = 3 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} \] **Answer:** \[ x = \frac{1}{2} \quad \text{and} \quad x = 1 \] --- If you have any further questions or need additional explanations, feel free to ask!