document.write( "Question 1102235: Log X + log(6x+1)=log2
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Algebra.Com's Answer #716890 by Boreal(15235)\"\" \"About 
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multiply when logs are added
\n" ); document.write( "log x(6x+1)= log 2
\n" ); document.write( "raise to 10th power to remove the logs
\n" ); document.write( "x(6x+1)=2
\n" ); document.write( "6x^2+x-2=0
\n" ); document.write( "(3x+2)(2x-1)=0
\n" ); document.write( "x=-2/3, 1/2 but logs of negative numbers don't exist
\n" ); document.write( "log 1/2 + log 4=log 2
\n" ); document.write( "that is log 2^(-1)+ log 2^2=log (2)^1 to check
\n" ); document.write( "x=1/2\r
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