document.write( "Question 858245: If a,b,c are three consecutive integers, prove that log(1+ac)=2logb. \n" ); document.write( "

Algebra.Com's Answer #517130 by josgarithmetic(39623) $\"\"$ $\"About$

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$\"log%28r%2C1%2Bac%29=2log%28r%2Cb%29\"$
\n" ); document.write( " $\"log%28r%2C1%2Bac%29=log%28r%2Cb%5E2%29\"$
\n" ); document.write( " $\"1%2Bac=b%5E2\"$
\n" ); document.write( " $\"b%5E2-1=ac\"$
\n" ); document.write( " $\"%28b-1%29%28b%2B1%29=ac\"$ , two factors for one member, two factors for the other member.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "You may assume by positions corresponding between left and right members, that
\n" ); document.write( "b-1=a and b+1=c. The number b, must be between a and c.
\n" ); document.write( "Using these same equations,
\n" ); document.write( "b=a+1 and b=c-1.
\n" ); document.write( "Seen starting at a,
\n" ); document.write( "a=b-1.
\n" ); document.write( "a+1=b-1+1=b. This shows that b is 1 unit more than a.
\n" ); document.write( "b+1=c as already found. This shows again, as a reminder, c is 1 unit more than b.\r
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\n" ); document.write( "\n" ); document.write( "This might not really count as a properly done proof; I only showed that $\"log%28r%2C1%2Bac%29=2log%28r%2Cb%29\"$ implies that a, b, and c are consecutive integers. What you asked for is the converse. \n" ); document.write( "

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