document.write( "Question 660718: Hello, I am stuck on a problem for homework and need some help. I worked it out but am not sure if I am right. Here is the problem and my answer. Can you check me please?\r
\n" ); document.write( "\n" ); document.write( "In(y-1)=1+In(3y+2)\r
\n" ); document.write( "\n" ); document.write( "I got an answer of y= (2e+1)/(1-3e)\r
\n" ); document.write( "\n" ); document.write( "Thank you very much for your help.\r
\n" ); document.write( "\n" ); document.write( "Chris \n" ); document.write( "
Algebra.Com's Answer #411329 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
In(y-1)=1+In(3y+2)\r
\n" ); document.write( "\n" ); document.write( "I got an answer of y= (2e+1)/(1-3e)
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\n" ); document.write( "I assume you mean ln, natural logs.
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\n" ); document.write( "ln(y-1)=1+ln(3y+2)
\n" ); document.write( "ln(y-1) = ln(e) + ln(3y+2)
\n" ); document.write( "ln(y-1) = ln(e*(3y+2))
\n" ); document.write( "y-1 = e*(3y+2)
\n" ); document.write( "y-1 = 3ey + 2e
\n" ); document.write( "y - 3ey = 2e + 1
\n" ); document.write( "y *(1 - 3e) = 2e + 1
\n" ); document.write( "y = (2e+1)/(1-3e)
\n" ); document.write( "-------------------
\n" ); document.write( "I concur, but y-1 --> a negative number.
\n" ); document.write( "You can't use a ln of a negative number.
\n" ); document.write( "--> no real solution.
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