52) \( -7 \cdot e^{7 m+2}=-29 \)

Solve the equation \( -7 \cdot e^{7m+2}=-29 \). Solve the equation by following steps: - step0: Solve for \(m\): \(-7e^{7m+2}=-29\) - step1: Change the sign: \(7e^{7m+2}=29\) - step2: Divide both sides: \(\frac{7e^{7m+2}}{7}=\frac{29}{7}\) - step3: Divide the numbers: \(e^{7m+2}=\frac{29}{7}\) - step4: Take the logarithm of both sides: \(\ln{\left(e^{7m+2}\right)}=\ln{\left(\frac{29}{7}\right)}\) - step5: Evaluate the logarithm: \(7m+2=\ln{\left(\frac{29}{7}\right)}\) - step6: Move the constant to the right side: \(7m=\ln{\left(\frac{29}{7}\right)}-2\) - step7: Divide both sides: \(\frac{7m}{7}=\frac{\ln{\left(\frac{29}{7}\right)}-2}{7}\) - step8: Divide the numbers: \(m=\frac{\ln{\left(\frac{29}{7}\right)}-2}{7}\) - step9: Simplify: \(m=\frac{\ln{\left(29\right)}-\ln{\left(7\right)}-2}{7}\) The solution to the equation \( -7 \cdot e^{7m+2}=-29 \) is \( m=\frac{\ln{(29)}-\ln{(7)}-2}{7} \).