(1) \( 9^{x+1}=27^{x} \) (2) \( 16^{x} \cdot 2^{x+1}=\sqrt[5]{4} \) (3) \( \sqrt[3]{9}=(\sqrt{3})^{2 x} \) (4) \( 8^{-x}-2 \cdot 4^{x-1}=0 \) (5) \( 3^{x+1}=\frac{\sqrt{3}}{3} \)

Let's solve each of the given exponential equations step by step. --- ### **Problem (1)** \[ 9^{x+1} = 27^{x} \] **Step 1:** Express both sides with the same base. - \( 9 = 3^2 \) - \( 27 = 3^3 \) So, the equation becomes: \[ (3^2)^{x+1} = (3^3)^x \] **Step 2:** Apply the power of a power rule \((a^m)^n = a^{mn}\): \[ 3^{2(x+1)} = 3^{3x} \] **Step 3:** Since the bases are the same, set the exponents equal to each other: \[ 2(x + 1) = 3x \] \[ 2x + 2 = 3x \] **Step 4:** Solve for \( x \): \[ 3x - 2x = 2 \] \[ x = 2 \] --- ### **Problem (2)** \[ 16^{x} \cdot 2^{x+1} = \sqrt[5]{4} \] **Step 1:** Express all terms with base 2. - \( 16 = 2^4 \) - \( 4 = 2^2 \) - \( \sqrt[5]{4} = 4^{1/5} = (2^2)^{1/5} = 2^{2/5} \) So, the equation becomes: \[ (2^4)^x \cdot 2^{x+1} = 2^{2/5} \] **Step 2:** Apply the power of a power rule: \[ 2^{4x} \cdot 2^{x+1} = 2^{2/5} \] **Step 3:** Combine the exponents on the left side: \[ 2^{4x + x + 1} = 2^{2/5} \] \[ 2^{5x + 1} = 2^{2/5} \] **Step 4:** Set the exponents equal to each other: \[ 5x + 1 = \frac{2}{5} \] **Step 5:** Solve for \( x \): \[ 5x = \frac{2}{5} - 1 \] \[ 5x = \frac{2}{5} - \frac{5}{5} \] \[ 5x = -\frac{3}{5} \] \[ x = -\frac{3}{25} \] --- ### **Problem (3)** \[ \sqrt[3]{9} = (\sqrt{3})^{2x} \] **Step 1:** Simplify both sides. - \( \sqrt[3]{9} = 9^{1/3} = (3^2)^{1/3} = 3^{2/3} \) - \( \sqrt{3} = 3^{1/2} \), so \( (\sqrt{3})^{2x} = (3^{1/2})^{2x} = 3^{x} \) So, the equation becomes: \[ 3^{2/3} = 3^{x} \] **Step 2:** Since the bases are the same, set the exponents equal: \[ \frac{2}{3} = x \] \[ x = \frac{2}{3} \] --- ### **Problem (4)** \[ 8^{-x} - 2 \cdot 4^{x-1} = 0 \] **Step 1:** Express all terms with base 2. - \( 8 = 2^3 \) - \( 4 = 2^2 \) So, the equation becomes: \[ (2^3)^{-x} - 2 \cdot (2^2)^{x-1} = 0 \] **Step 2:** Apply the power of a power rule: \[ 2^{-3x} - 2 \cdot 2^{2(x-1)} = 0 \] \[ 2^{-3x} - 2 \cdot 2^{2x - 2} = 0 \] **Step 3:** Simplify the second term: \[ 2 \cdot 2^{2x - 2} = 2^{1} \cdot 2^{2x - 2} = 2^{2x - 1} \] So, the equation becomes: \[ 2^{-3x} - 2^{2x - 1} = 0 \] **Step 4:** Set the exponents equal since the bases are the same: \[ -3x = 2x - 1 \] **Step 5:** Solve for \( x \): \[ -3x - 2x = -1 \] \[ -5x = -1 \] \[ x = \frac{1}{5} \] \[ x = \frac{1}{5} \] --- ### **Problem (5)** \[ 3^{x+1} = \frac{\sqrt{3}}{3} \] **Step 1:** Simplify the right side. - \( \sqrt{3} = 3^{1/2} \) So, \[ \frac{\sqrt{3}}{3} = \frac{3^{1/2}}{3^{1}} = 3^{1/2 - 1} = 3^{-1/2} \] **Step 2:** The equation becomes: \[ 3^{x+1} = 3^{-1/2} \] **Step 3:** Since the bases are the same, set the exponents equal: \[ x + 1 = -\frac{1}{2} \] **Step 4:** Solve for \( x \): \[ x = -\frac{1}{2} - 1 \] \[ x = -\frac{3}{2} \] \[ x = -\frac{3}{2} \] --- ### **Summary of Solutions:** 1. **Problem (1):** \( x = 2 \) 2. **Problem (2):** \( x = -\frac{3}{25} \) 3. **Problem (3):** \( x = \frac{2}{3} \) 4. **Problem (4):** \( x = \frac{1}{5} \) 5. **Problem (5):** \( x = -\frac{3}{2} \)