SOLUTION: if log(a+b+c)=loga+logb+logc then prove that log(2a/1-a^2 + 2b/1-b^2 + 2c/1-c^2)= log 2a/1-a^2 + log 2b/1-b^2 + log 2c/1-c^2

Question 1043663: if log(a+b+c)=loga+logb+logc then prove that log(2a/1-a^2 + 2b/1-b^2 + 2c/1-c^2)= log 2a/1-a^2 + log 2b/1-b^2 + log 2c/1-c^2
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
We wish to show that implies that
.
For the result to have meaning, it must be that 0 < a,b,c < 1, which can be determined by solving the inequalities
, , and .
With this in mind, it is enough to show that
a+b+c = abc ===> . <---- Why?

Now on to the proof:

===>, since a+b+c = abc,
===>
===>
===>
<===>
<===> after a little rearrangement of terms.
Since a+b+c = abc, we get

===>
<===>
<===>
<===> , after dividing both sides by

And that's it...

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