#### Answer

The solution is
$$\lim_{x\to\infty}\frac{\ln(3x+5)}{\ln(7x+3)+1}=1.$$

#### Work Step by Step

To solve this limit follow the steps below. "LR" will stand for "Apply L'Hopital's rule."
$$\lim_{x\to\infty}\frac{\ln(3x+5)}{\ln(7x+3)+1}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln(3x+5))'}{(\ln(7x+3)+1)'}=\lim_{x\to\infty}\frac{\frac{1}{3x+5}(3x+5)'}{\frac{1}{7x+3}(7x+3)'}=\lim_{x\to\infty}\frac{\frac{3}{3x+5}}{\frac{7}{7x+3}}=\lim_{x\to\infty}\frac{3(7x+3)}{7(3x+5)}=\lim_{x\to\infty}\frac{3x\left(7+\frac{3}{x}\right)}{7x\left(3+\frac{5}{x}\right)}=\lim_{x\to\infty}\frac{3\left(7+\frac{3}{x}\right)}{7\left(3+\frac{5}{x}\right)}=\left[\frac{3\left(7+\frac{3}{\infty}\right)}{7\left(3+\frac{5}{\infty}\right)}\right]=\left[\frac{3(7+0)}{7(3+0)}\right]=1.$$