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This log problem is throwing me off, out of nowhere

  • Thread starter Jurrasic
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  • #1
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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?
 

Answers and Replies

  • #2
Mentallic
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Think about the formula ln(a)-ln(b) = ln(a/b) :wink:
 
  • #3
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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.
 
Last edited:
  • #4
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Think about the formula ln(a)-ln(b) = ln(a/b) :wink:
Thats exactly what is above where the answer is wrong , where you get -1 on one side
 
  • #5
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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
 
  • #6
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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. :wink:
 
  • #7
Mentallic
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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.
[tex]\ln\left(x+1\right)\neq \ln\left(x-1\right)[/tex]
so you can't just take away the ln(...) from both sides to get 0=-1, use the formula I suggested and try again.
 
  • #8
SammyS
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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.

[itex]\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}[/itex]

See what you can do with that.
 
  • #9
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You can use each side of the equation,
ln(x+1)-1 = ln(x-1)​
as an exponent with a base of e.

[itex]\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}[/itex]

See what you can do with that.
Alternatively, it might be simpler to write the first equation above as
[itex]ln(x + 1) - ln(x - 1) = 1[/itex]

and then use a property of logs to write the left side as ln of a single expression. Then exponentiate both sides.
 
  • #10
SammyS
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Alternatively, it might be simpler to write the first equation above as
[itex]ln(x + 1) - ln(x - 1) = 1[/itex]

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.
 
  • #11
Mentallic
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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? :smile:
 
  • #12
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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.
 
  • #13
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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).
 
  • #14
SammyS
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You can use each side of the equation,
ln(x+1)-1 = ln(x-1)​
as an exponent with a base of e.

[itex]\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}[/itex]

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

[itex]\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}[/itex]

is equivalent to:

[itex]\displaystyle e^{\ln(x+1)}e^{-1}=e^{\ln(x-1)}[/itex]

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

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