## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to\infty}\frac{\log_2 x}{\log_3 x}=\frac{\ln 3}{\ln 2}.$$
We will use the logarithmic rule for base change which says $$\log_{b}a=\frac{\ln a}{\ln b}.$$ This gives $$\log_{2}x=\frac{\ln x}{\ln 2},\quad \log_{3}x=\frac{\ln x}{\ln 3}.$$ Putting this into the limit we get $$\lim_{x\to\infty}\frac{\log_{2}x}{\log_{3}x}=\lim_{x\to\infty}\frac{\frac{\ln x}{\ln 2}}{\frac{\ln x}{\ln 3}}=\lim_{x\to\infty}\frac{\ln 3}{\ln 2}=\frac{\ln 3}{\ln 2}.$$