Question 858245: If a,b,c are three consecutive integers, prove that log(1+ac)=2logb.
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39623)   (Show Source):
You can put this solution on YOUR website! 
 , 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 implies that a, b, and c are consecutive integers. What you asked for is the converse.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! If a,b,c are three consecutive integers, prove that
log(1+ac)=2logb.
log(1+ac)=logb^2
b^2=1+ac
..
let a=1st consecutive integer=a
a+1=2nd consecutive integer=b
a+2=3rd consecutive integer=c
..
(a+1)^2=1+a(a+2)
a^2+2a+1=1+a^2+2a
verified: log(1+ac)=2logb
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