document.write( "Question 1072286: If a^x=bc;b^y=ca;c^z=ab then prove x/(x+1) + y/(y+1) + z/(z+1) =2 \n" ); document.write( "
Algebra.Com's Answer #687204 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
\"a%5Ex\"\"%22%22=%22%22\"\"bc\";\"b%5Ey\"\"%22%22=%22%22\"\"ca\";\"c%5Ez\"\"%22%22=%22%22\"\"ab\" then prove \"x%2F%28x%2B1%29+%2B+y%2F%28y%2B1%29+%2B+z%2F%28z%2B1%29\"\"%22%22=%22%22\"\"2\"
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document.write( "To avoid doing essentially the same thing three times, \r\n" );
document.write( "I'll just do it once with the letters p,q,r,s and use it as\r\n" );
document.write( "a formula for each of the fractions:\r\n" );
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document.write( "\"p%5Eq\"\"%22%22=%22%22\"\"r%2As\"\r\n" );
document.write( "\"log%28%28p%5Eq%29%29\"\"%22%22=%22%22\"\"log%28%28r%2As%29%29\"\r\n" );
document.write( "\"q%2Alog%28%28p%29%29\"\"%22%22=%22%22\"\"log%28%28r%29%29%2Blog%28%28s%29%29\"\r\n" );
document.write( "\"q\"\"%22%22=%22%22\"\"%28log%28%28r%29%29%2Blog%28%28s%29%29%29%2F%28log%28%28p%29%29%29\"\r\n" );
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document.write( "Therefore,\r\n" );
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document.write( "\"%22%22=%22%22\"\r\n" );
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document.write( "and therefore, using the above as a formula,\r\n" );
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document.write( "\"x%2F%28x%2B1%29+%2B+y%2F%28y%2B1%29+%2B+z%2F%28z%2B1%29\"\"%22%22=%22%22\"\r\n" );
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document.write( "\"%22%22=%22%22\"\"%22%22=%22%22\"\"%22%22=%22%22\"\"2\"\r\n" );
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document.write( "Edwin

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