document.write( "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 \n" ); document.write( "
Algebra.Com's Answer #660462 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
We wish to show that \"log%28%28a%2Bb%2Bc%29%29=loga%2Blogb%2Blogc\" implies that\r
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\n" ); document.write( "\n" ); document.write( "For the result to have meaning, it must be that 0 < a,b,c < 1, which can be determined by solving the inequalities
\n" ); document.write( "\"2a%2F%281-a%5E2%29%3E0\", \"2b%2F%281-b%5E2%29%3E0\", and \"2c%2F%281-c%5E2%29%3E0\".\r
\n" ); document.write( "\n" ); document.write( "With this in mind, it is enough to show that \r
\n" ); document.write( "\n" ); document.write( "a+b+c = abc ===> . <---- Why?\r
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\n" ); document.write( "\n" ); document.write( "Now on to the proof:\r
\n" ); document.write( "\n" ); document.write( "\"-%28ab%2Bac%2Bbc%29abc+%2B+%28ab%2Bbc%2Bac%29abc+=+0\"\r
\n" ); document.write( "\n" ); document.write( "===>\"-%28ab%2Bac%2Bbc%29%28a%2Bb%2Bc%29+%2B+%28ab%2Bbc%2Bac%29abc+=+0\", since a+b+c = abc,\r
\n" ); document.write( "\n" ); document.write( "===> \"-ab%28a%2Bb%2Bc%29+-+ac%28a%2Bb%2Bc%29-+bc%28a%2Bb%2Bc%29+%2B+%28ab%2Bbc%2Bac%29abc+=+0\"\r
\n" ); document.write( "\n" ); document.write( "===> \r
\n" ); document.write( "\n" ); document.write( "===> \"-ab%28a%2Bb%29+-ac%28a%2Bc%29-bc%28b%2Bc%29+%2B+%28ab%2Bbc%2Bac%29abc+=+3abc\"\r
\n" ); document.write( "\n" ); document.write( "<===> \"-a%5E2b+-+ab%5E2+-+a%5E2c+-+ac%5E2+-+b%5E2c-bc%5E2+%2B+%28ab%2Bbc%2Bac%29abc+=+3abc\"\r
\n" ); document.write( "\n" ); document.write( "<===> \"-+ab%5E2-+ac%5E2+-+a%5E2b+-+bc%5E2++-+a%5E2c+-+b%5E2c++%2B+%28ab%2Bbc%2Bac%29abc+=+3abc\" after a little rearrangement of terms.\r
\n" ); document.write( "\n" ); document.write( "Since a+b+c = abc, we get \r
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\n" ); document.write( "\n" ); document.write( "===> \r
\n" ); document.write( "\n" ); document.write( "<===> \r
\n" ); document.write( "\n" ); document.write( "<===> \r
\n" ); document.write( "\n" ); document.write( "<===> , after dividing both sides by \"+%281-a%5E2%29%281-b%5E2%29%281-c%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "And that's it...\r
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