Prove that $\log(n + 10) = Θ(\log n)$

Question

Would $\log(n + 10) = Θ(\log n)$?

I was having trouble with this because I wasn't sure if I could get rid of the $10$ inside the log as a lower order term.

Answer 1

Hint :

prove that $$\log n < \log(n+10) < 2\log n$$

for some $n > N$. Consider that $\log(n+10) = \log n + \log(1+\frac{10}{n})$.