document.write( "Question 1099231: logb(x-1) + logb3 = logbx\r
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document.write( "I thought answer was -1.5 , that seems to be wrong.
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document.write( "First I applied the product rule, so
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document.write( "log_b(x+1) x log_b(3) = log_b(x)
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document.write( "log_b(3x+3) =lob_b(X)
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document.write( "log_b(3x+3)-log_b(x)=0
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document.write( "Raise all the terms to power of log b
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document.write( "3x+3-x=0
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document.write( "2x+3=0
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document.write( "2x=-3
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document.write( "x=-3/2 \n" );
document.write( "
Algebra.Com's Answer #713667 by Alan3354(69443) You can put this solution on YOUR website! logb(x-1) + logb3 = logbx\r \n" ); document.write( "\n" ); document.write( "I thought answer was -1.5 , that seems to be wrong. \n" ); document.write( "First I applied the product rule, so \n" ); document.write( "log_b(x+1) x log_b(3) = log_b(x) \n" ); document.write( "log_b(3x+3) =lob_b(X) \n" ); document.write( "log_b(3x+3)-log_b(x)=0 \n" ); document.write( "Raise all the terms to power of log b \n" ); document.write( "3x+3-x=0 \n" ); document.write( "2x+3=0 \n" ); document.write( "2x=-3 \n" ); document.write( "x=-3/2 \n" ); document.write( "================= \n" ); document.write( "Your method and calculations are correct, but that gives \n" ); document.write( "log(-5/2) + log(3) = log(-3/2) \n" ); document.write( "Logs of negative numbers are not allowed. \n" ); document.write( "--> no solution \n" ); document.write( " |