Show that $\frac{\log_aN-\log_bN}{\log_bN-\log_cN}=\frac{\log_aN}{\log_cN}$ Show that $$\dfrac{\log_aN-\log_bN}{\log_bN-\log_cN}=\dfrac{\log_aN}{\log_cN}$$ where $a,b$ and $c$ are positive and are consecutive terms of а geometric sequence, $a\ne1,b\ne1,c\ne1,N>0,N\ne1$.
$a,b$ and $c$ are consecutive terms of a geometric sequence if and only if $b^2=ac, b=\sqrt{ac}$. Then the LHS is $$\dfrac{\log_aN-\log_\sqrt{ac}N}{\log_\sqrt{ac}N-\log_cN}=\dfrac{\log_aN-2\log_{ac}N}{2\log_{ac}N-\log_cN}$$ This seems useless. What can we do? What is the intuition that will lead to the solution? Thank you!
 A: $$\dfrac{\log_aN-\log_bN}{\log_bN-\log_cN}-\dfrac{\log_aN}{\log_cN}$$
$$=\dfrac{\frac{1}{\log_Na}-\frac{1}{\log_Nb}}{\frac{1}{\log_Nb}-\frac{1}{\log_Nc}}-\frac{\frac{1}{\log_Na}}{\frac{1}{\log_Nc}}$$
$$=\dfrac{\log c\log b-\log a \log c}{\log a\log c-\log a \log b}-\frac{\log c}{\log a}$$
(where all logs are base $N$)
$$=\frac{\log c}{\log a}\left[\frac{\log b-\log a}{\log c - \log b}-1\right]$$
But $$\log b -\log a =\log c-\log b = \log r,$$
where $r$ is the common ratio. Therefore the expression is zero as required.
A: Well, using the fact that:
$$\log_\alpha\beta=\frac{\ln\beta}{\ln\alpha}\tag1$$
We get:
$$\frac{\log_\text{a}\text{N}-\log_\text{b}\text{N}}{\log_\text{b}\text{N}-\log_\text{c}\text{N}}=\frac{\frac{\ln\text{N}}{\ln\text{a}}-\frac{\ln\text{N}}{\ln\text{b}}}{\frac{\ln\text{N}}{\ln\text{b}}-\frac{\ln\text{N}}{\ln\text{c}}}=\frac{\frac{\ln\text{N}\ln\text{b}-\ln\text{N}\ln\text{a}}{\ln\text{a}\ln\text{b}}}{\frac{\ln\text{N}\ln\text{c}-\ln\text{N}\ln\text{b}}{\ln\text{b}\ln\text{c}}}=$$
$$\frac{\ln\text{N}\ln\text{b}-\ln\text{N}\ln\text{a}}{\ln\text{a}\ln\text{b}}\cdot\frac{\ln\text{b}\ln\text{c}}{\ln\text{N}\ln\text{c}-\ln\text{N}\ln\text{b}}=\frac{\ln\text{b}-\ln\text{a}}{\ln\text{a}}\cdot\frac{\ln\text{c}}{\ln\text{c}-\ln\text{b}}\tag2$$
A: From where you left -
$ \displaystyle \frac{\log_aN-2\log_{ac}N}{2\log_{ac}N-\log_cN} = \frac{\log_aN-2(\log_aN)(\log_{ac}a)}{2(\log_cN)(\log_{ac}c)-\log_cN}$
$ \displaystyle  = \frac{\log_aN}{\log_cN} \cdot \frac{1 - 2\log_{ac}a}{2\log_{ac}c-1}$
Now note that by dividing by $\log_{ac}c$,
$ \displaystyle \frac{1 - 2\log_{ac}a}{2\log_{ac}c-1} = \frac{\log_{c}{ac} - 2\log_{c}a}{2 - \log_{c}{ac}} = \frac{1 - \log_ca}{1 - \log_ca} = 1$
