## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to\infty}(\log_2 x -\log_3 x)=\infty.$$
To solve this limit we will use the change of basis rule to switch to the basis of $2$ in the second term: $$\log_3 x=\frac{\log_2 x}{\log_2 3}=\frac{1}{\log_2 3}\log_2 x.$$ Now we will the other logarithmic rule, namely that $a\log_b c=\log_b c^a$ which applied to this problem gives $$\frac{1}{\log_2 3}\log_2 x=\log_2x^{\frac{1}{\log_2 3}}$$ Finally, applying the difference-of-the-logarithms rule we get that $$\log_2 x-\log_2 x^{\frac{1}{\log_2 3}}=\log_2\frac{x}{x^\frac{1}{\log_2 3}}=\log_2 x^{1-\frac{1}{\log_2 3}}.$$ Putting this into the limit we have $$\lim_{x\to\infty}(\log_2 x-\log_3 x)=\lim_{x\to\infty}(\log_2 x-\log_2x^{\frac{1}{\log_2 3}})=\lim_{x\to\infty}\log_2 x^{1-\frac{1}{\log_2 3}}=\log_2\lim_{x\to\infty}x^{1-\frac{1}{\log_2 3}}=[\log_2 \infty]=\infty.$$ In the last step we used the fact that $\log_2 3>1$ so $1/\log_2 3<1$ and that finally gives $1-1/\log_23>0.$ When $x$ tends to $\infty$ then every power of $x$ that is greater then zero also tends to $\infty$ and thats why we concluded that $x^{1-\frac{1}{\log_2 3}}\to\infty.$