document.write( "Question 6831: Since a logarithm is an exponent, how do you think the log property
\n" ); document.write( "logb(xy) = logb (x) + logb (y) is related to the exponent property
\n" ); document.write( " (b^m)(b^n) = b^(m+n)?
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Algebra.Com's Answer #3821 by prince_abubu(198)\"\" \"About 
You can put this solution on YOUR website!
First let \"+J+=+log%28b%2Cx%29+%2B+log%28b%2Cy%29+\"\r
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\n" ); document.write( "\n" ); document.write( "So then \"+log%28b%2Cxy%29+=+J+\". The equivalent exponent for that log equation is \"+b%5EJ+=+xy+\".\r
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\n" ); document.write( "\n" ); document.write( "Now, let's put back the logs in place of the J (un-substitute):\r
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\n" ); document.write( "\n" ); document.write( "\"+b%5E%28log%28b%2Cx%29+%2B+log%28b%2Cy%29%29+=+xy+\".\r
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\n" ); document.write( "\n" ); document.write( "Remember the rule that says \"+b%5E%28m+%2B+n%29+=+b%5Em%2Ab%5En+\"\r
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\n" ); document.write( "\n" ); document.write( "Applying that above exponential rule, \"+b%5Elog%28b%2Cx%29%2Ab%5Elog%28b%2Cy%29+=+xy+\".\r
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\n" ); document.write( "\n" ); document.write( "Remember the rule \"+b%5Elog%28b%2Cm%29+=+m+\". In this case, \"+b%5Elog%28b%2Cx%29+=+x+\" and \"+b%5Elog%28b%2Cy%29+=+y+\". So we're all cool. \r
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