In the inequality

$0.85^x \lt \frac{2}{9}$

to solve for $x$ one way to solve is to take the log of both sides: (or take the log of both sides with a base greater than 1)

$\log{}0.85^x \lt \log{}\frac{2}{9}$

$x\log{}0.85 \lt \log{}\frac{2}{9}$

$x \lt \frac{log{}\frac{2}{9}}{\log{}0.85}$

But since $\log{}0.85$ and $\log{}\frac{2}{9}$ are both negative numbers, the final answer is positive but the dividing of $\log{}0.85$ causes the inequality symbol to flip. So it actually is:

$x \gt \frac{log{}\frac{2}{9}}{\log{}0.85}$

$x \gt \ 0.9.2548...$

Another way is to solve by taking the log base 0.85 of both sides: (or take the log of both sides with a base less than 1, greater than 0)

$\log_{0.85}0.85^x \lt \log_{0.85}\frac{2}{9}$

Since $\log_{0.85}0.85^x$ is just $x$,

$x \lt \log_{0.85}\frac{2}{9}$

$x \lt 9.2548...$

The two methods both yield the same numerical value, but the inequality is flipped in the first method due to the division of the negative number and not in the second method.

Why is this? Both methods seem valid to me. And is there a more correct method?


1 Answer 1


Inequalities are

  • preserved (or not flipped) when you apply a strictly increasing function; but
  • reversed (or flipped) when you apply a strictly decreasing function.

For $b>1$, $\log_b$ is a strictly increasing function. (So in particular, $\log_{\mathrm e}$ or $\log$ or $\ln$ is a strictly increasing function and so everything you did in your first method is correct.)

For $0<b<1$, $\log_b$ is a strictly decreasing function. (So when you applied $\log_{0.85}$, you made an error--you should have flipped the inequality.)

  • $\begingroup$ So I guess the rule that "inequalities are flipped when multiplying by a negative" is too specific, and what you provided is a more general rule. Thanks! $\endgroup$
    – Oongabu
    Mar 30 at 5:38
  • $\begingroup$ And where can I find a proof for this rule of when the inequality is preserved? $\endgroup$
    – Oongabu
    Mar 30 at 5:40
  • 1
    $\begingroup$ @Oongabu: By definition of a strictly increasing function, if $a<b$, then $f(a)<f(b)$. $\endgroup$
    – user986614
    Mar 30 at 5:48

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