5. \( \frac{2}{3+\log (x)}+\frac{1}{7-\log (x)}=\frac{3}{5} \) 6. \( \frac{5}{\log _{2}(x+1)+1}-\frac{2}{3 \log _{2}(x+1)-3}=1 \) 7. \( \ln (3+x)-\ln (2)=\ln (5)-\ln (4-x) \) 8. \( \log _{x}(2 x+2)+\log _{x}(x-1)=2 \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\log_{x}{\left(2x+2\right)}+\log_{x}{\left(x-1\right)}=2\) - step1: Find the domain: \(\log_{x}{\left(2x+2\right)}+\log_{x}{\left(x-1\right)}=2,x>1\) - step2: Transform the expression: \(\log_{x}{\left(\left(2x+2\right)\left(x-1\right)\right)}=2\) - step3: Convert the logarithm into exponential form: \(\left(2x+2\right)\left(x-1\right)=x^{2}\) - step4: Expand the expression: \(2x^{2}-2=x^{2}\) - step5: Move the expression to the left side: \(2x^{2}-2-x^{2}=0\) - step6: Subtract the terms: \(x^{2}-2=0\) - step7: Move the constant to the right side: \(x^{2}=0+2\) - step8: Remove 0: \(x^{2}=2\) - step9: Simplify the expression: \(x=\pm \sqrt{2}\) - step10: Separate into possible cases: \(\begin{align}&x=\sqrt{2}\\&x=-\sqrt{2}\end{align}\) - step11: Check if the solution is in the defined range: \(\begin{align}&x=\sqrt{2}\\&x=-\sqrt{2}\end{align},x>1\) - step12: Find the intersection: \(x=\sqrt{2}\) Solve the equation \( \ln (3+x)-\ln (2)=\ln (5)-\ln (4-x) \). Solve the equation by following steps: - step0: Solve for \(x\): \(\ln{\left(3+x\right)}-\ln{\left(2\right)}=\ln{\left(5\right)}-\ln{\left(4-x\right)}\) - step1: Find the domain: \(\ln{\left(3+x\right)}-\ln{\left(2\right)}=\ln{\left(5\right)}-\ln{\left(4-x\right)},-3