1

I saw the following in a book:

$$\log_{1/2}x \geq \log_{1/3}x$$

$$\Rightarrow \log_2x \geq \log_3x$$

Now, none of the properties I know deals with fractional bases. What's the justification behind this step?

Community
  • 1
ankush981
  • 2,033

2 Answers2

2

$$\log_{1/k}x = \frac{\ln x}{\ln \frac{1}{k}} = -\frac{\ln x}{\ln k} = -\log_{k}x$$ so $$\log_{1/2}x \geq \log_{1/3}x \Rightarrow \log_2x \leq \log_3x$$

Clement C.
  • 67,323
  • But the inequality is not matching. Does that mean it's a mistake in the book? – ankush981 Aug 05 '13 at 16:08
  • Either in the book or in my answer. However, trying with $x=1/10$, it looks like the book is wrong. – Clement C. Aug 05 '13 at 16:09
  • Your answer seems fine. – Javier Aug 05 '13 at 16:10
  • I'm not so sure. At $x = 1/10, \log_{1/2}x = 3.32$ and $\log_{1/3}x = 2.1$. Thus, $\log_{1/2}x > \log_{1/3}x$. Also, if we consider $\log_x2 = -0.3$ and $\log_x3 = -0.47$, we see that $\log_x2 > \log_x3$. Or have I made a mistake somewhere? – ankush981 Aug 05 '13 at 16:17
  • Is there a property that says $-\log_ab = \log_ba$? I don't think so, but it would make life so much more easy. – ankush981 Aug 05 '13 at 16:18
  • Well, as a matter of fact, for any $x> 1$ the first inequality is false; so the implication trivially holds for $x > 1$ (and is also true for $x=1$). So the statement you are trying to prove is correct for $x\geq 1$ (but actually "empty"). – Clement C. Aug 05 '13 at 16:18
  • @dotslash: for $x < 1$, $\log_{1/2}x \geq \log_{1/3}x$ but $\log_{2}x < \log_{3}x$. – Clement C. Aug 05 '13 at 16:20
  • @dotslash The RHS is reciprocal of what you have in comment 5 – Harish Kayarohanam Aug 05 '13 at 16:20
  • @ClementC. Actually the question revolves around establishing that $0 < x < 1$. Would that make the textbook correct? – ankush981 Aug 05 '13 at 16:24
  • No. The implication is wrong for $x\in(0,1)$, and correct for $x\in[1,\infty)$. – Clement C. Aug 05 '13 at 17:24
0

First, let me start by saying that the second inequality in your original post is pointing in the wrong direction. Use your calculator with a fixed $x$ and check.

Let's think about what $\log_{1/2}x$ means. If $y = \log_{1/2}x$ then, by definition, $(1/2)^y = x$. Notice that $(1/2)^y \equiv 1/2^y$ and, assuming $x \neq 0$, we have $(1/2)^y = x \iff 2^y=1/x$. Going back to the definition of logs, if $2^y = 1/x$ then $$y = \log_2(1/x) \equiv \log_21-\log_2x \equiv - \log_2x$$ We started with $y = \log_{1/2}x$ and ended up with $y = -\log_2x$. It follows that $\log_{1/2}x\equiv-\log_2x$. Likewise, $\log_{1/3}x\equiv-\log_3x$. Hence: $$\log_{1/2}x \ge \log_{1/3}x \iff -\log_2x \ge -\log_3x \iff \log_2x \le \log_3x$$

Fly by Night
  • 32,272