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

Question

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

Is the inequality right in the second line? I'd say that $0.8^n>\frac{1}{13}$ and, therefore, that all subsequent inequalities should also be reversed. — Patricio, Nov 08 '18 at 11:16

Eff · Accepted Answer · 2018-11-08T11:20:18.080

1

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 $\log x <0$.

edited Nov 08 '18 at 11:20

answered Nov 08 '18 at 11:12

Eff

13,304

Simple as that! Thanks! Will mark accepted when I can. While you're here, any chance of letting me know why my title is not formatting correctly? :) – Henry Cooper Nov 08 '18 at 11:14
@HenryCooper You accidentally had an ) where you needed an }, for the fraction that is. I've corrected it for you. – Eff Nov 08 '18 at 11:16
1

Hero, thanks, much appreciated! – Henry Cooper Nov 08 '18 at 11:18

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

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