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

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The discussion revolves around solving the equation ln((x-1)/(x-3))=2. The user initially derives x = 2/(e^2-1) but is confused by the lecturer's solution of x = (3e^2-1)/(e^2-1). A key step involves multiplying both sides of the equation by (x-3) to rearrange it into the form x(e^2 - 1) = 3e^2 - 1. The correct manipulation leads to the lecturer's solution, which approximates to 3.313035285. The thread concludes with a reminder to continue discussions in a linked thread to avoid duplicates.
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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
 
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blackfriars said:
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 $x = \dfrac{3e^2-1}{e^2-1} \approx 3.313035...$.
 
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
 
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
 
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