Solve for \( x: 4^{2 x+1}=8^{3 x+2} \) \( x= \) \( 0-\frac{3}{4} \) \( 0-\frac{4}{5} \)

To solve the equation \( 4^{2x+1} = 8^{3x+2} \), we first express both sides in terms of powers of 2: We know that: - \( 4 = 2^2 \), so \( 4^{2x+1} = (2^2)^{2x+1} = 2^{4x + 2} \) - \( 8 = 2^3 \), so \( 8^{3x+2} = (2^3)^{3x+2} = 2^{9x + 6} \) Now, we have: \[ 2^{4x + 2} = 2^{9x + 6} \] Since the bases are the same, we can set the exponents equal to each other: \[ 4x + 2 = 9x + 6 \] Now, let's solve for \( x \): 1. Subtract \( 4x \) from both sides: \[ 2 = 5x + 6 \] 2. Subtract \( 6 \) from both sides: \[ -4 = 5x \] 3. Divide by 5: \[ x = -\frac{4}{5} \] So the solution is \( x = -\frac{4}{5} \). To check: Substituting \( x = -\frac{4}{5} \) back into the original equation: - LHS: \[ 4^{2(-\frac{4}{5})+1} = 4^{-\frac{8}{5}+1} = 4^{-\frac{3}{5}} = (2^2)^{-\frac{3}{5}} = 2^{-\frac{6}{5}} \] - RHS: \[ 8^{3(-\frac{4}{5})+2} = 8^{-\frac{12}{5}+2} = 8^{-\frac{12}{5}+\frac{10}{5}} = 8^{-\frac{2}{5}} = (2^3)^{-\frac{2}{5}} = 2^{-\frac{6}{5}} \] Since both sides equal \( 2^{-\frac{6}{5}} \), the solution \( x = -\frac{4}{5} \) is confirmed.