# This log problem is throwing me off, out of nowhere

Question is: solve for x
ln(x+1)-1 = ln(x-1)

tried to do it TWO different ways, and neither is the right way, check this out:
first attempt:
since LN is on both sides , it goes away and you just have variables, that way sucks because then you get 0=-1 , wrong.
second attempt:
divide both sides by ln(x+1)
that leaves -1 on one side and a bunch of nonsensical stuff on the other side.
WHAT TO DO?

Mentallic
Homework Helper
Think about the formula ln(a)-ln(b) = ln(a/b) Question is: solve for x
ln(x+1)-1 = ln(x-1)

since LN is on both sides , it goes away and you just have variables, that way sucks because then you get 0=-1 , wrong.

Your first assumption was not correct. Yet your subsequent reasoning was logical and you yourself noticed that the conclusion did not make sense.

You do not have ln(....) = In(...)
but you have In(...) + 1 = In(...).

Last edited:
Think about the formula ln(a)-ln(b) = ln(a/b) Thats exactly what is above where the answer is wrong , where you get -1 on one side

Your first assumption was not correct. Yet your subsequent reasoning was logical and you yourself noticed that the conclusion did not make sense.

You do not have ln(....) = In(...)
but you have In(...) + 1 = In(...).

already did that and the answer was wrong, see above. answers not needed, but steps and how to

Thats exactly what is above where the answer is wrong , where you get -1 on one side

-1=ln(x-1)-ln(x+1).
Now try applying what Mentallic said. Mentallic
Homework Helper
already did that and the answer was wrong, see above. answers not needed, but steps and how to

Is this what you're referring to?

first attempt:
since LN is on both sides , it goes away and you just have variables, that way sucks because then you get 0=-1 , wrong.

$$\ln\left(x+1\right)\neq \ln\left(x-1\right)$$
so you can't just take away the ln(...) from both sides to get 0=-1, use the formula I suggested and try again.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Question is: solve for x
ln(x+1)-1 = ln(x-1)

tried to do it TWO different ways, and neither is the right way, check this out:
first attempt:
since LN is on both sides , it goes away and you just have variables, that way sucks because then you get 0=-1 , wrong.
second attempt:
divide both sides by ln(x+1)
that leaves -1 on one side and a bunch of nonsensical stuff on the other side.
WHAT TO DO?

You can use each side of the equation,
ln(x+1)-1 = ln(x-1)​
as an exponent with a base of e.

$\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}$

See what you can do with that.

Mark44
Mentor
You can use each side of the equation,
ln(x+1)-1 = ln(x-1)​
as an exponent with a base of e.

$\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}$

See what you can do with that.

Alternatively, it might be simpler to write the first equation above as
$ln(x + 1) - ln(x - 1) = 1$

and then use a property of logs to write the left side as ln of a single expression. Then exponentiate both sides.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Alternatively, it might be simpler to write the first equation above as
$ln(x + 1) - ln(x - 1) = 1$

and then use a property of logs to write the left side as ln of a single expression. Then exponentials both sides.

Yes, Mark, I agree, but OP seemed to be unable to do this.

Mentallic
Homework Helper
Yes, Mark, I agree, but OP seemed to be unable to do this.

But I"m failing to see how I could do anything with your suggestion other than to go back to where the question started. Can I get a hint? The simplest way to do it is to get it in the form
In(x + 1) - In(x -1) = 1
and then use In(a) - In(b) = In(a/b) as the other posters suggested.

Mark44
Mentor
The simplest way to do it is to get it in the form
In(x + 1) - In(x -1) = 1
and then use In(a) - In(b) = In(a/b) as the other posters suggested.
There is no "In" function. It is ln (ell en), an abbreviation for natural logarithm (logarithm naturalis, in Latin).

SammyS
Staff Emeritus
Homework Helper
Gold Member
You can use each side of the equation,
ln(x+1)-1 = ln(x-1)​
as an exponent with a base of e.

$\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}$

See what you can do with that.
Have I been asked what this suggestion was good for?

$\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}$

is equivalent to:

$\displaystyle e^{\ln(x+1)}e^{-1}=e^{\ln(x-1)}$

I hope OP can simplify $\displaystyle e^{\ln(x+1)}\text{ and }e^{\ln(x-1)}$