<s>[INST] 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!
 [/INST] 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$
</s>