10. \( \log _{3} x+\log _{3}(2 x+3)=2 \)

To solve the equation \( \log _{3} x+\log _{3}(2 x+3)=2 \), we can start by using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \). So, we can combine the logarithmic terms as follows: \[ \log_3 x + \log_3 (2x + 3) = \log_3 (x(2x + 3)) \] This allows us to rewrite the equation as: \[ \log_3 (x(2x + 3)) = 2 \] Next, we convert the logarithmic equation into its exponential form: \[ x(2x + 3) = 3^2 \] \[ x(2x + 3) = 9 \] Now we'll expand and rearrange the equation: \[ 2x^2 + 3x - 9 = 0 \] This is a quadratic equation, and we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = 3, c = -9 \): \[ b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-9) = 9 + 72 = 81 \] Thus, the solutions for \( x \) are: \[ x = \frac{-3 \pm \sqrt{81}}{2 \cdot 2} = \frac{-3 \pm 9}{4} \] Calculating the two possible values: 1. \( x = \frac{6}{4} = \frac{3}{2} \) 2. \( x = \frac{-12}{4} = -3 \) Since \( x \) must be positive (as logarithms of non-positive numbers are undefined), we discard \( x = -3 \). Thus, the solution to the equation is: \[ \boxed{\frac{3}{2}} \]