Solve each exponential equation. Round your answer to the nearest hundredth. \( \begin{array}{ll}\text { a. } 9^{n+2}+3=52 & \text { b. } \frac{7^{2-2}}{5}=2\end{array} \)

To solve these exponential equations, let’s tackle them one at a time! For equation \( a \): Start with the equation \( 9^{n+2} + 3 = 52 \). Subtract 3 from both sides to get \( 9^{n+2} = 49 \). Notice that \( 9 \) can be expressed as \( 3^2 \). Thus, we can rewrite the equation as \( (3^2)^{n+2} = 49 \), or \( 3^{2(n+2)} = 7^2 \). This leads us to \( 2(n+2) = 2 \log_3(7) \). Next, divide each side by 2 to isolate \( n+2 \): \( n+2 = \log_3(7) \). Lastly, subtract 2 to find \( n \): \( n = \log_3(7) - 2 \). Using a calculator for \( \log_3(7) \) gives approximately \( 1.77124 \). Thus, rounding gives \( n \approx -0.23 \). For equation \( b \): The equation \( \frac{7^{2-2}}{5} = 2 \) simplifies, since \( 7^{2-2} = 7^0 = 1 \). Thus, we have \( \frac{1}{5} = 2 \), which clearly isn't true. Therefore, this equation has no solution since a fraction cannot equal a whole number in this case.