Question: (6) $9^{2x+1}=27$

Solution

Rewrite \( 9 \) and \( 27 \) as powers of \( 3 \): \[ 9 = 3^2 \] \[ 27 = 3^3 \] Rewrite the original equation using base \( 3 \): \[ (3^2)^{2x+1} = 3^3 \] Simplify the left side: \[ 3^{4x + 2} = 3^3 \] Set the exponents equal: \[ 4x + 2 = 3 \] Solve for \( x \): \[ 4x = 1 \] \[ x = \frac{1}{4} \]