Solve ln(x-1)/x-3=2 | Step-by-Step Guide

[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=x(e^2-1)-3
but the lecturer gave a solution of

3e^2-1/e^2-1 = 3.313035285 how do i get to this

the solution i got was this 2/e^2-1 =0.3130352855

 

[MarkFL]

Okay, we are given:

$$\ln\left(x-\frac{1}{x}-3\right)=2$$

Converting from logarithmic to exponential form, we have:

$$x-\frac{1}{x}-3=e^2$$

Multiply through by $x$:

$$x^2-1-3x=e^2x$$

Arrange in standard quadratic form:

$$x^2-\left(e^2+3\right)x-1=0$$

Applying the quadratic formula, we obtain:

$$x=\frac{e^2+3\pm\sqrt{\left(e^2+3\right)^2+4}}{2}=\frac{e^2+3\pm\sqrt{e^4+6e^2+13}}{2}$$

This agrees with:

W|A - ln(x-1/x-3)=2

 

[blackfriars]

sorry this is what i meant , i wrote the equation the wrong way

this is the correct way

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

 

[Greg]

$$\ln\left(\frac{x-1}{x-3}\right)=2$$

$$\frac{x-1}{x-3}=e^2$$

$$x-1=xe^2-3e^2$$

$$x-xe^2=1-3e^2$$

$$x(1-e^2)=1-3e^2$$

$$x=\frac{1-3e^2}{1-e^2}$$

$$x=\frac{3e^2-1}{e^2-1}$$

 

[blackfriars]

hi , in your 3rd line of work where did the 2nd (e^2) come from
thanks

 

[Greg]

$$\frac{x-1}{x-3}=e^2$$

$$x-1=e^2(x-3)$$

$$x-1=xe^2-3e^2$$

 

[blackfriars]

thank you i could not see that brilliant