Question 858245
{{{log(r,1+ac)=2log(r,b)}}}
{{{log(r,1+ac)=log(r,b^2)}}}
{{{1+ac=b^2}}}
{{{b^2-1=ac}}}
{{{(b-1)(b+1)=ac}}}, two factors for one member, two factors for the other member.


Notice that each factor on the left is either 1 unit more than be or 1 unit less than b.  This means that a and c are two units difference from each other.


You may assume by positions corresponding between left and right members, that 
b-1=a and b+1=c.  The number b, must be between a and c.
Using these same equations,
b=a+1 and b=c-1.
Seen starting at a,
a=b-1.
a+1=b-1+1=b.  This shows that b is 1 unit more than a.
b+1=c as already found.  This shows again, as a reminder, c is 1 unit more than b.


This might not really count as a properly done proof; I only showed that {{{log(r,1+ac)=2log(r,b)}}} implies that a, b, and c are consecutive integers.  What you asked for is the converse.