document.write( "Question 283616: Solve
\n" ); document.write( "Log6(x+3) + Log6(x-2) = 1
\n" ); document.write( "* 6 is the base for both logs
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Algebra.Com's Answer #205831 by user_dude2008(1862)\"\" \"About 
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Log6 (x+3) + Log6 (x-2) = 1\r
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\n" ); document.write( "\n" ); document.write( "Log6 [(x+3)(x-2)] = 1\r
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\n" ); document.write( "\n" ); document.write( "(x+3)(x-2) = 6^1\r
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\n" ); document.write( "\n" ); document.write( "(x+3)(x-2) = 6\r
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\n" ); document.write( "\n" ); document.write( "x^2+x-6 = 6\r
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\n" ); document.write( "\n" ); document.write( "x^2+x-12=0\r
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\n" ); document.write( "\n" ); document.write( "(x+4)(x-3)=0\r
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\n" ); document.write( "\n" ); document.write( "x+4=0 or x-3=0\r
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\n" ); document.write( "\n" ); document.write( "x=-4 or x=3\r
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\n" ); document.write( "\n" ); document.write( "x=-4 is extraneous\r
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\n" ); document.write( "\n" ); document.write( "Answer: Only solution is x=3
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