document.write( "Question 63579: Write as a single logarithm: log 12 + log (x-2) - log 2.
\n" ); document.write( "I thought that I might have to turn it into exponetial form first but Im not sure.
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Algebra.Com's Answer #44305 by praseenakos@yahoo.com(507)\"\" \"About 
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ANSWER:\r
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\n" ); document.write( "\n" ); document.write( "Write as a single logarithm: log 12 + log (x-2) - log 2.
\n" ); document.write( "I thought that I might have to turn it into exponetial form first but Im not sure.\r
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\n" ); document.write( "\n" ); document.write( "LAWS OF LOGARITHM:\r
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\n" ); document.write( "\n" ); document.write( " logA + logB = log AB\r
\n" ); document.write( "\n" ); document.write( "log A - log B = log (A/B)\r
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\n" ); document.write( "\n" ); document.write( "So, log 12 + log (x-2) = log {12 * (x-2)}\r
\n" ); document.write( "\n" ); document.write( " = log {12(x-2)}\r
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\n" ); document.write( "\n" ); document.write( " so, log 12 + log (x-2) - log 2 = log {12(x-2)} - log 2\r
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\n" ); document.write( "\n" ); document.write( " = log {12(x-2)/2}\r
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\n" ); document.write( "\n" ); document.write( " = log {6(x-2)}\r
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\n" ); document.write( "\n" ); document.write( " = log {6x- 12}\r
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\n" ); document.write( "\n" ); document.write( "Hope you understood.\r
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\n" ); document.write( "\n" ); document.write( "Regards,\r
\n" ); document.write( "\n" ); document.write( "praseenakos@yahoo.co.in\r
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