# How to know if $\log_78 > \log_89$ without using a calculator?

I realize that I lack any numerical intuition for logarithms. For example, when comparing two logarithms like $\log_78$ and $\log_89$, I have to use the change-of-base formula and calculate the values with a calculator in order to determine which of the two is larger.

Can anyone demonstrate an algebraic/analytic method to find out which of the two is larger without using a calculator?

• $\log_x(x+1)=\log(x+1)/\log(x)$ is decreasing for $x>1$ but I don't know how to prove it...
– lhf
May 3, 2012 at 10:50
• @lhf: $\frac{\log(x + 2) / \log (x + 1)}{\log(x + 1) / \log(x)} = \frac{\log(x + 2) \log(x)}{\log(x + 1)^2}$. Now $\frac{(x + 2)x}{(x + 1)^2} < 1$ (rectangle areas), and since $\log$ is concave, the same must hold for when you apply $\log$ to each factor. May 3, 2012 at 12:22

$\log_7 8 = 1 + \log_7 (8 / 7) > 1 + \log_7 (9/8) > 1 + \log_8(9/8) = \log_8 9$

• brilliant! Thanks. May 3, 2012 at 11:23

The derivative of $f(x)=\frac{\log(x+1)}{\log(x)}$ has the same sign as $\frac{\log x}{x+1}-\frac{\log(x+1)}{x}$ which is negative if $x>1$ since $x\mapsto x\log{x}$ is increasing.

Of course, this method does not apply for arbitrary 7,8,9. For example $\log_35$ and $\log_23$ are quite close and proving that one is bigger than the other is not so easy. The only elegant way I know is some kind of trick. (Enjoy this entertaining exercise! Spoiler below.)

Prove that $\log_35 < \frac32 < \log_23$.

• I don't get why the two expressions have the same sign. But actually I'm looking for a general method to compare the sizes of two logarithms. So the numbers 7,8,9 in the question are actually arbitrary. May 3, 2012 at 11:03
• @dragoncharmer: $f'(x)\big(\log x\big)^2=\frac{\log x}{x+1}-\frac{\log(x+1)}x$, and $\big(\log x\big)^2>0$ for $x\ne 1$. May 3, 2012 at 11:11
• interesting exercise. it seems to me wallace's method fails here. May 3, 2012 at 11:35
• btw in the exercise, 1/2 should be 3/2. May 3, 2012 at 11:43
• @dragoncharmer: oh, thanks.
May 3, 2012 at 12:07

Alternate solution:

$$\log_78 > \log_89 \Leftrightarrow \frac{1}{\log_87} > \log_89 \Leftrightarrow 1> \log_87 \log_89$$

Now, by AM-GM

$$\log_87 \log_89 \leq (\frac{\log_87+ \log_89}{2})^2=(\frac{\log_863}{2})^2< (\frac{\log_864}{2})^2=1$$

In general If $ab < c^2$ and $\log_b c >0$ you have

$$\log_ca \log_cb \leq (\frac{\log_ca+ \log_cb}{2})^2< 1$$

and thus

$$\log_ca < \log_bc \,.$$

Consider the function $f(x)=\log_x(x+1)$

This is a decreasing function and hence $\log_n(n+1) >\log_m(m+1)$ for $n < m$ and hence $\log_78 > \log_89$.

$f(x)$ is decreasing because, $f'(x)$ comes out to be negative,

$$f(x)=\frac{\ln(x+1)}{\ln(x)}$$

$$f '(x)=\frac{1}{(x+1)\ln x} + \ln(x+1).\frac{-1}{{(\ln x)}^2}.\frac{1}{x}$$

$$f '(x)=\frac{1}{\ln x}\left(\frac{1}{x+1}-\frac{f(x)}{x}\right)$$

and since $f(x)<1$ for $x \in \mathbb R^+$, we have $f'(x)<0$.