MHB Natural logs solve ln⁡((x-1)/(x-3))=2

[blackfriars]

hi , hope someone can help as i can't get past a certain step
the natural logs is the problem
ln⁡((x-1)/(x-3))=2
i can get to this point here -1 = e_x^2-x-3
-1+3=x(ⅇ^2-1)
2 = x(ⅇ^2-1)
2/((ⅇ^2-1) )=x((ⅇ^2-1)/(ⅇ^2-1))
X = 2/(ⅇ^2-1)
the solution i got was this x= (2/(ⅇ^2-1)) → 0.3130352855

but the lecturer gave a solution of
(3ⅇ^2-1)/(ⅇ^2-1) = 3.313035285 how do i get to this

 

[Opalg]

hi , hope someone can help as i can't get past a certain step
the natural logs is the problem
ln⁡((x-1)/(x-3))=2
i can get to this point here -1 = e_x^2-x-3
-1+3=x(ⅇ^2-1)
2 = x(ⅇ^2-1)
2/((ⅇ^2-1) )=x((ⅇ^2-1)/(ⅇ^2-1))
X = 2/(ⅇ^2-1)
the solution i got was this x= (2/(ⅇ^2-1)) → 0.3130352855

but the lecturer gave a solution of
(3ⅇ^2-1)/(ⅇ^2-1) = 3.313035285 how do i get to this

If $\ln\left(\frac{x-1}{x-3}\right) = 2$ then $\frac{x-1}{x-3} = e^2.$ Multiply both sides by $x-3$ to get $x-1 = (x-3)e^2.$ Then rearrange that as $x(e^2 - 1) = 3e^2 - 1$. That gives $x = \dfrac{3e^2-1}{e^2-1} \approx 3.313035...$.

 

[blackfriars]

hi sorry for the questions but i cannot transpose the formula to make x the subject could you show the workings for making x the subject
thanks

 

[Greg]

Same topic and a working found http://mathhelpboards.com/pre-algebra-algebra-2/logs-22167-new.html. Thread closed - please continue discussion in linked thread.

blackfriars, please do not post duplicate topics; thanks. :D