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

1. Nov 4, 2011

### 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?

2. Nov 4, 2011

### Mentallic

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

3. Nov 4, 2011

### grzz

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: Nov 4, 2011
4. Nov 4, 2011

### Jurrasic

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

5. Nov 4, 2011

### Jurrasic

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

6. Nov 4, 2011

### Saitama

-1=ln(x-1)-ln(x+1).
Now try applying what Mentallic said.

7. Nov 4, 2011

### Mentallic

Is this what you're referring to?

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

8. Nov 4, 2011

### SammyS

Staff Emeritus
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.

9. Nov 5, 2011

### Staff: Mentor

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.

10. Nov 5, 2011

### SammyS

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

11. Nov 5, 2011

### Mentallic

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?

12. Nov 5, 2011

### 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.

13. Nov 5, 2011

### Staff: Mentor

There is no "In" function. It is ln (ell en), an abbreviation for natural logarithm (logarithm naturalis, in Latin).

14. Nov 5, 2011

### SammyS

Staff Emeritus
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 $\displaystyle e^{\ln(x+1)}\text{ and }e^{\ln(x-1)}$