This log problem is throwing me off, out of nowhere

[Jurrasic]

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]

Think about the formula ln(a)-ln(b) = ln(a/b)

 

[grzz]

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(...).

Follow the advice of Mentallic.

 

[Jurrasic]

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

 

[Jurrasic]

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(...).

Follow the advice of Mentallic.

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

 

[Saitama]

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]

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]

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]

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]

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]

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?

 

[grzz]

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]

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]

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)}