# Natural Logarithms

1. Sep 18, 2007

### physstudent1

1. The problem statement, all variables and given/known data
ln(x^2 + 1 ) -3lnx=ln2

Solve for x.

2. Relevant equations

3. The attempt at a solution

I used laws of logarithms to simplify it down to -x^2(x-1)=1 . I don't think this is the answer I'm under the impression you need x = for the answer. First I brought the 3 up to a exponent and then put everything e^ to get rid of the ln's. Could someone please point me in the right direction.

2. Sep 18, 2007

### D H

Staff Emeritus
Show your work on how you arrived at this result and we can help you further (hint: its not correct).

3. Sep 18, 2007

### physstudent1

sure thing;

I started with ln(x^2+1)-3lnx = ln2

I brought the 3 up so its

ln(x^2+1)-lnx^3=ln2
then I did

e^ln(x^2+1)-e^ln(x^3)=e^ln2

and got

x^2+1 - x^3 = 2
-1 from both sides
x^2-x^3 = 1

-x^2(-1+x)=1

4. Sep 18, 2007

### D H

Staff Emeritus
OK so far. You're mistake is here:
Can you see what you did wrong or do you need a bit more help?

5. Sep 18, 2007

### physstudent1

ln((x^2+1)/(x^3)) = ln2

before doing the e^ ?

6. Sep 18, 2007

### D H

Staff Emeritus
That works. So does using [itex]e^{a+b} = e^ae^b[/tex]. You did the equivalent of [itex]e^{a+b} = e^a+e^b[/tex], which is incorrect.

7. Sep 18, 2007

### physstudent1

I'm not sure how to use that to help

but doing it from the way I stated

I put

e^ln((x^2+1)/x^3) = e^ln2

(x^2+1)/x^3 = 2 If this is correct I'm not sure how to do the algebra to get a single x

Last edited: Sep 18, 2007
8. Sep 18, 2007

### D H

Staff Emeritus
Multiply both sides by x3. You won't get a single x (its a cubic equation, after all), but only of the three solutions is a real number.

9. Sep 18, 2007

### Hurkyl

Staff Emeritus
Before you started working with exponentials and logarithms, didn't you spend a lot of time solving equations just like (x^2+1)/x^3 = 2?

10. Sep 18, 2007

### physstudent1

yes I can get it to a cubic expression of 2x^3-x^2 -1 i just didn't htink this was right I thought it was looking for a x = would this be considered the correct answeR?

11. Sep 18, 2007

### Hurkyl

Staff Emeritus
I assume you meant to say you can derive the cubic equation
2x^3-x^2 -1 = 0.​

Again I ask, haven't you spent a lot of time learning how to solve equations exactly like this?

12. Sep 18, 2007

### physstudent1

the only thing I remeber about solving equations like this is using P's / Q's and testing the 0's however I don't remeber what the P's over Q's were is this the correct direction?

Edit:

Is the answer x=1 ( and yes I have in precalc and algebra 2 but those were like 2 and 3 years ago and I am a bit rusty on this)

Last edited: Sep 18, 2007
13. Sep 18, 2007

Plug it in and see what happens.

14. Sep 18, 2007

### Hurkyl

Staff Emeritus
In terms of studying, math resembles foreign languages -- you cannot forget what you've already learned if you want to progress! You shouldn't be afraid to pull out your old textbooks to refresh your memory when necessary.

Anyways, x=1 is, in fact, one of the solutions to that cubic equation. Now that you know one solution, do you remember what to do to find the rest of the solutions?

(And, incidentally, you always need to check that the answers you derive actually satisfy the original equation: some common algebraic techniques can introduce fake solutions)

15. Sep 18, 2007

### physstudent1

yes I plugged it in and it worked out :); can you keep using synthetic division that is what I used to find the first solution