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?
$\log_7 8 = 1 + \log_7 (8 / 7) > 1 + \log_7 (9/8) > 1 + \log_8(9/8) = \log_8 9$
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 $\log_35 < \frac32 < \log_23$.
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 $f'(x)<0$.
Let $f(x):=\log_x(x+1)$, with $x>1$. Then $$f(x)=\log_x x+\log_x\left(1+\frac1x\right)=1+\frac{\ln(1+1/x)}{\ln x}.$$ Now, as $x$ increases, $1+1/x$ strictly decreases, as does its logarithm; at the same time, the denominator $\ln x$ increases. Hence $f$ is a strictly decreasing function.It follows that $$\log_78>\log_89.$$
Given $x>1$, we set $$\tag{1} y=1-\frac{1}{x}, \quad z=1+\frac{1}{x} $$ so that $$\tag{2} x-1=xy, \quad x+1=xz, \quad yz=1-\frac{1}{x^2}. $$ Notice that $$\tag{3} 0<y, yz <1, \quad 1< z <2 $$ Now \begin{eqnarray}\tag{4} \log(x-1)&=&\log(xy)=\log(x)+\log(y)\cr \log(x+1)&=&\log(xz)=\log(x)+\log(z)\cr \log(x-1)\log(x+1)&=&\log^2(x)+[\log(y)+\log(z)]\log((x)+\log(y)\log(z)\cr &=&\log^2(x)+\log(yz)\log(x)+\log(y)\log(z) \end{eqnarray} Thanks to the fact that $x>1$ together with (3) we have $$\tag{5} \max(\log(y), \log(yz) )<0, \quad \min(\log(x), \log(z))>0 $$ Combining (5) to the last equation in (4), we obtain $$ \log(x-1)\log(x+1)<\log^2(x) $$ or equivalently $$\tag{6} \log_{x-1}(x)> \log_x(x+1) \quad \forall x>1 $$ Setting $x=8$, we get the desired inequality.
