SOLUTION: 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? In(

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?
In(y-1)=1+In(3y+2)
I got an answer of y= (2e+1)/(1-3e)
Thank you very much for your help.
Chris
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
In(y-1)=1+In(3y+2)
I got an answer of y= (2e+1)/(1-3e)
-----------
I assume you mean ln, natural logs.
------
ln(y-1)=1+ln(3y+2)
ln(y-1) = ln(e) + ln(3y+2)
ln(y-1) = ln(e*(3y+2))
y-1 = e*(3y+2)
y-1 = 3ey + 2e
y - 3ey = 2e + 1
y *(1 - 3e) = 2e + 1
y = (2e+1)/(1-3e)
-------------------
I concur, but y-1 --> a negative number.
You can't use a ln of a negative number.
--> no real solution.

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