Solve the equation. \[ 4^{x+1}=8^{2 x-3} \]

To solve the equation \( 4^{x+1} = 8^{2x-3} \), we can first express both sides of the equation in terms of powers of 2. We know that: - \( 4 = 2^2 \), so \( 4^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x+2} \). - \( 8 = 2^3 \), so \( 8^{2x-3} = (2^3)^{2x-3} = 2^{3(2x-3)} = 2^{6x-9} \). Now we can rewrite the original equation: \[ 2^{2x+2} = 2^{6x-9}. \] Since the bases are the same, we can set the exponents equal to each other: \[ 2x + 2 = 6x - 9. \] Now, let's solve for \( x \): 1. Rearranging gives us: \[ 2 + 9 = 6x - 2x, \] which simplifies to: \[ 11 = 4x. \] 2. Dividing both sides by 4: \[ x = \frac{11}{4}. \] Thus, the solution to the equation is: \[ \boxed{\frac{11}{4}}. \]