# Math Help - Natural Log problem, please can you help?

1. ## Natural Log problem, please can you help?

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
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!

2. = ln y - ln( -y + 1)

= ln y - ln (1/y) + ln 1 therefore
This step is incorrect. I'm not sure I would approach the problem this way. I'd go like this:

x=5+ln(y/(y-1))

x - 5 = ln(y/(y-1))

e^(x-5) = y/(y-1)

...

3. Thanks for your quick response
I will give this way a try and let you know if I am still struggling.