document.write( "Question 20730: Can anyone help solve this?\r
\n" ); document.write( "\n" ); document.write( "1 - logx = log(3x-1)\r
\n" ); document.write( "\n" ); document.write( "I tried:
\n" ); document.write( "1 - logx = log 3x - log 1
\n" ); document.write( "1 = logx + log 3x - log 1
\n" ); document.write( "1 = log4x - log 1\r
\n" ); document.write( "\n" ); document.write( "I also tried:
\n" ); document.write( "1 - logx = log(3x-1)
\n" ); document.write( "divide both sides by log
\n" ); document.write( "1-x = (3x-1)
\n" ); document.write( "2 - x = 3x
\n" ); document.write( "2 = 4x
\n" ); document.write( "1/2 = x\r
\n" ); document.write( "\n" ); document.write( "Or:
\n" ); document.write( "1 - logx = log(3x-1)
\n" ); document.write( "1 - 10 = log3x - log
\n" ); document.write( "-9 = 3logx - log
\n" ); document.write( "-9 = 3(10) - log
\n" ); document.write( "-9 = 30 - log
\n" ); document.write( "-39 = -log\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The teacher's answer is 2
\n" ); document.write( "Can anyone help?\r
\n" ); document.write( "\n" ); document.write( "Thanks,
\n" ); document.write( "Sandy
\n" ); document.write( "

Algebra.Com's Answer #9966 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
\"1-logx+=+log%28%283x-1%29%29\" Add logx to both sides of the equation.
\n" ); document.write( "\"1+=+log%28%283x-1%29%29+%2B+logx\" Apply the product rule for logarithms.
\n" ); document.write( "\"1+=+log%28%28%283x-1%29%2A%28x%29%29%29\"
\n" ); document.write( "\"1+=+log%28%283x%5E2-x%29%29\" Recall that: If \"log%28a%29+=+1\", then \"a+=+10\" because: \"log%2810%29=1\", therefore:
\n" ); document.write( "\"%283x%5E2-x%29+=+10\" Subtract 10 from both sides.
\n" ); document.write( "\"3x%5E2-x-10=+0\" Solve this quadratic equation by factoring.
\n" ); document.write( "\"%283x%2B5%29%28x-2%29+=+0\" Apply the zero products principle.
\n" ); document.write( "\"%283x%2B5%29+=+0\" and/or \"%28x-2%29+=+0\"
\n" ); document.write( "If \"3x%2B5+=+0\" then \"3x+=+-5\" and \"x+=+-5%2F3\"
\n" ); document.write( "If \"x-2+=+0\" then \"x+=+2\"\r
\n" ); document.write( "\n" ); document.write( "The roots are:
\n" ); document.write( "\"x+=+2\"
\n" ); document.write( "\"x+=+%28-5%2F3%29\"\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "1) x = 2
\n" ); document.write( "\"1-log%282%29+=+log%28%283%282%29-1%29%29\"
\n" ); document.write( "\"1-0.30103+=+log%285%29\"
\n" ); document.write( "\"0.69897+=+0.69897\"\r
\n" ); document.write( "\n" ); document.write( "2) \"x+=+-5%2F3\"
\n" ); document.write( "1-log((-5/3)) = 0.77815..., -1.36437...
\n" ); document.write( "log(3(-5/3)-1) = 0.77815..., 1.36437...
\n" ); document.write( "\"x+=+%28-5%2F3%29\" is not a valid root.\r
\n" ); document.write( "\n" ); document.write( "Answer is x = 2
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