Q:

Why is $n>\frac{-\log 13}{\log 0.8}=n \log0.8<-\log13$

Not sure if I'm missing something obvious here but in my text book, dealing with a logarithmic inequality, the $<$ is seemingly switched arbitrarily:
$$
  1-0.8^n < \frac{12}{13}
$$
$$
  0.8^n < \frac{1}{13}
$$
$$
  n \log0.8 < -\log13
$$
$$
  n > \frac{-\log13}{\log0.8}
$$
I saw this type of rearranging earlier in the book and thought it might be a typo but seeing it a second time confirms there's something I'm not understanding. Why has the sign changed although we have simply divided both sides by $\log0.8$?
Thanks in advance

A:

Because $\log 0.8$ is negative, and dividing by a negative number reverses inequalities (see for example this question).

More generally, you have that $\log 1 = 0$ because $e^{0} = 1$. So when $0<x<1$ you have that $\log x <0$.

