Show that $\log_{10}(x^2 + 1)$ and $\log_2 x$ have the same order of growth [closed]

There are three things you need to know here. One is the rule

$$\log_a x^y = y\log_a x,$$

telling you that squares or other powers in the logarithm make no difference to speed. The other is

$$\log ab = \log a + log b$$ which you surely know. The last rule is the formula for the change of base in logarithms, that is $$log_b x = \frac{\log_a x}{\log_a b}$$

With the two rules, I can write

$$\log_{10} (x^2 + 1) = \log_{10}\left(x^2\left(1 + \frac{1}{x^2}\right)\right) = \log_{10}x^2 + \log_{10}\left(1 + \frac{1}{x^2}\right)=\\=2\log_{10} x + log_{10}\left(1 + \frac{1}{x^2}\right) = 2\frac{\log_2 x}{\log_2 10} + log_{10}\left(1 + \frac{1}{x^2}\right)=\\=\frac{2}{\log_210}\log_2 x + log_{10}\left(1 + \frac{1}{x^2}\right).$$

The conclusion is simple. Since $1+1/x^2$ is a bounded function (its limit is even $0$, so it does not affect growth at all, much less the order of growth), and $\frac{2}{\log_210}$ is a constant, the functions have the same order of growth.

Using that $(\log_{a})'=\cdots$ and L'Hopital in $$\lim_{x\to\infty}{\log_{10} (x^2+1)\over\log_2 x}=\cdots$$

Hint:

First ask yourself whether $\log_{10}\left(x^{2}+1\right)$ and $\log_{10}x^{2}$ have the same order of growth. If so then it is enough to compare $\log_{2}x=\frac{\log_{10}x}{\log_{10}2}$ and $\log_{10}x^{2}=2\log_{10}x$.