This step is incorrect. I'm not sure I would approach the problem this way. I'd go like this:= ln y - ln( -y + 1)
= ln y - ln (1/y) + ln 1 therefore
x=5+ln(y/(y-1))
x - 5 = ln(y/(y-1))
e^(x-5) = y/(y-1)
...
Hello
I am new to this forum so therefore please accept my apologies if I have not posted this problem to the correct forum.
x=5+ln(y/(y-1))
I am trying to break out of the brackets to the above problem
Please can someone confirm that I have gone about it correctly.
Also can you confirm that I have solved for Y correctly. I am not confident with this at all.
1) we know that natural log ln(a/b) = ln a - ln b therefore
2) ln(y/(1-y)) = ln y - ln(1-y)
= ln y - ln( -y + 1)
= ln y - ln (1/y) + ln 1 therefore
x = ln y - ln 1 - ln y + ln 1 + 5
3) Ok now to get Y on its own.
ln y - ln 1 - ln y + ln 1 + 5 = x
ln y - ln 1 + ln 1 - ln y + ln 1 + 5 = x + ln 1
ln y - ln y + ln 1 + 5 = x + ln 1
ln y - ln y + ln y + ln 1 + 5 = x + ln 1 + ln y
ln y + ln 1 + 5 = x + ln 1 + ln y
ln y + ln 1 - ln 1 + 5 = x + ln 1 + ln y - ln 1
ln y + 5 - 5 = x + ln 1 + ln y - ln 1 - 5
ln y = x + ln 1 + ln y - ln 1 - 5
ln y / ln y = x + ln 1 + ln y - ln 1 - 5 / ln y
so 1 or y = x + ln 1 + ln y - ln 1 - 5 / ln y
This is how I am seeing it at the moment, I know that this cannot be right as the quotient equation for Y would be undefined as the denominator is equal to zero!
Thanks for your time
Hi,
Assuming you were correct when you wrote ln y - ln 1 - ln y + ln 1 + 5 = x, why didn't you say or think that ln y - ln y cancel out, ie equals 0 and that ln 1 - ln 1 also equals 0? So the left side is just 5 and so x = 5. You can't solve for Y since there are no y's since they canceled out.
Your main problem is, as the previous poster stated so well, is that ln(-y + 1) can't be broken down. Yes ln (ab) = ln (a) + ln (b) and ln (a/b) = ln (a) - ln (b) but ln (a +/- b) is NOT ln (a) +/- ln (b). Now solve for ln(y/(y-1)) in x=5+ln(y/(y-1)).
Good luck.