# Why does taking logarithms on both sides of $0<r<s$ reverse the inequality for logarithms with base $a$, $0<a<1$?

Why does taking logarithms on both sides of $0<r<s$ reverse the inequality for logarithms with base $a$, $0<a<1$?

I would like some intuition on why this works. I tried graphing $\log_{0.5}(x)$ on Desmos, for example, and if the graph were true this would be evident from the graph, but the graph seems wrong because I don't get why as $x\to0^+$, $y\to \infty$ since $0.5^x$ should be $\le 0.5$ where $0<x<1$.

• $\log_{0.5}(x) = \ln(x)/\ln(0.5)$ is decreasing because $\ln(0.5)$ is negative. And $0.5^x > 0.5$ for $0<x<1$. – Martin R Nov 12 '17 at 2:50

Because logs with bases less than $1$ are decreasing, not increasing functions. As an example, we have $8 \lt 16$, but $\log_{0.5} 8=-3 \gt \log_{0.5}16=-4$. It is like multiplying by a negative number, which is a decreasing function and reverses the inequality.
• We might add that (or even begin with the thought that) exponentials of such an $a$ are decreasing functions, $0.5^2>0.5^3$ for example. – Mathemagical Nov 12 '17 at 3:00
Because $\log_a(b) =\dfrac{\log_c b}{\log_c a}$ for any $a, b, c > 0$.
The usual thing is to choose $c=e$ or $c=10$; the key point is that in both cases $c > 1$.
Therefore, if $b > 1$ and $0 < a < 1$ then $\log_c b > 0$ and $\log_c a < 0$ so that $\log_a b < 0$ and the usual inequalities are reversed.