TITLE: How to know if $\log_78 > \log_89$ without using a calculator?
QUESTION [11 upvotes]: 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?

REPLY [2 votes]: 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 $f'(x)<0$.