58) \( -10 \ln (3 x+1)+6=-24 \)

Solve the equation \( -10 \ln (3x+1)+6=-24 \). Solve the equation by following steps: - step0: Solve for \(x\): \(-10\ln{\left(3x+1\right)}+6=-24\) - step1: Find the domain: \(-10\ln{\left(3x+1\right)}+6=-24,x>-\frac{1}{3}\) - step2: Move the expression to the left side: \(-10\ln{\left(3x+1\right)}+6-\left(-24\right)=0\) - step3: Subtract the numbers: \(-10\ln{\left(3x+1\right)}+30=0\) - step4: Solve using substitution: \(-10t+30=0\) - step5: Move the constant to the right side: \(-10t=0-30\) - step6: Remove 0: \(-10t=-30\) - step7: Change the signs: \(10t=30\) - step8: Divide both sides: \(\frac{10t}{10}=\frac{30}{10}\) - step9: Divide the numbers: \(t=3\) - step10: Substitute back: \(\ln{\left(3x+1\right)}=3\) - step11: Convert the logarithm into exponential form: \(3x+1=e^{3}\) - step12: Move the constant to the right side: \(3x=e^{3}-1\) - step13: Divide both sides: \(\frac{3x}{3}=\frac{e^{3}-1}{3}\) - step14: Divide the numbers: \(x=\frac{e^{3}-1}{3}\) - step15: Check if the solution is in the defined range: \(x=\frac{e^{3}-1}{3},x>-\frac{1}{3}\) - step16: Find the intersection: \(x=\frac{e^{3}-1}{3}\) The solution to the equation \( -10 \ln (3x+1)+6=-24 \) is \( x=\frac{e^{3}-1}{3} \).