If `((log)_a N)/((log)_c N)=((log)_a N-(log)_b N)/((log)_b N-(log)_c N),w h e r eN >0a n dN!=1, a , b , c >0` and not equal to 1, then prove that `b^2=a c`

Explanation

`log_a^N/log_c^N=(log_a^N-log_b^N)/(log_a^N-log_c^N)`

`(logN/loga)/(logN/logc)=((logN/loga)-(logN/logb))/((logN/logb)-(logN/logc))`

`logc/loga=((logb-loga)/(logalogb))/((logc-logb)/(logblogc))`

`logc/loga=((logb-loga)logc)/((logc-loga)loga)`

`logc-logb=logb-loga`

`loga+logc=2logb`

`logac=logb^2`

`b^2=ac`.