Solve for \( x \) in \( 9^{x+1}+3^{2 x+1}=36 \)

To solve the equation \( 9^{x+1} + 3^{2x+1} = 36 \), we start by rewriting \( 9 \) and \( 3 \) in terms of their base: 1. We know that \( 9 = 3^2 \), so we can rewrite \( 9^{x+1} \) as \( (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} \). 2. The term \( 3^{2x+1} \) remains as is. Now substituting these into the equation: \[ 3^{2x + 2} + 3^{2x + 1} = 36 \] Next, we can factor out the common term \( 3^{2x + 1} \): \[ 3^{2x + 1} (3 + 1) = 36 \] \[ 3^{2x + 1} \cdot 4 = 36 \] Now, divide both sides by \( 4 \): \[ 3^{2x + 1} = 9 \] Since \( 9 = 3^2 \), we equate the exponents: \[ 2x + 1 = 2 \] Now, solving for \( x \): \[ 2x = 2 - 1 \] \[ 2x = 1 \] \[ x = \frac{1}{2} \] Thus, the solution for \( x \) is \[ \boxed{\frac{1}{2}}. \]