Solve logarithmic equation \( \rightarrow \log _{3}\left(b^{2}+15\right)=2 \)

To solve the equation \( \log_{3}(b^{2}+15) = 2 \), we can eliminate the logarithm by rewriting the equation in exponential form. This means we convert the logarithmic equation into its equivalent exponential form: \[ b^{2} + 15 = 3^{2} \] Calculating \( 3^{2} \) gives us 9: \[ b^{2} + 15 = 9 \] Now, we can solve for \( b^{2} \) by isolating it: \[ b^{2} = 9 - 15 \] This simplifies to: \[ b^{2} = -6 \] Since the square of a real number cannot be negative, we conclude that there are no real solutions for \( b \). Therefore, the equation \( \log_{3}(b^{2}+15) = 2 \) has no solution in the real number set. If you were looking for complex solutions, then we can state: \[ b = \pm i\sqrt{6} \] So to sum up, the solution is \( b = i\sqrt{6} \) and \( b = -i\sqrt{6} \).