# Calculus Natural Log

1. Sep 10, 2010

### Ericalvusa91

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

2. Relevant equations
Solve for x

3. The attempt at a solution
Well I put the ln on the left side
ln(3x+1)+ln(x) = 3

Then I combine them
ln ((3x+1)(x)) = 3

So I take e

e^ln(x(3x+1)) = e^3

I get x(3x+1) = e^3

So I divide the x

(3x+1)/(x) = (e^3)/(x)

Multiply x by both sides
3x+1 = e^3

Subtract the 1
3x = (e^3)-1

Divide by 3
x = ((e^3)-1)/3

Can someone tell me if I made a mistake or did it totally wrong ?

2. Sep 10, 2010

### jgens

When you divide by x, you should get 3x+1 = e3x-1. This is different than what you show and it should have an impact on the final answer.

3. Sep 10, 2010

### Ericalvusa91

So I did what you said.
3x+1 = e^3 x^-1

3x-x^(-1) +1 =e^3

factor out x(3-1^(-1))=e^3

Divide both sides by 3-1^(-1)

ans: (e^3)/ 2 But I still get it wrong

4. Sep 10, 2010

### jgens

How did you get this? It doesn't follow from the equality you had before this.

5. Sep 10, 2010

### Ericalvusa91

I subtracted the x^-1

6. Sep 10, 2010

### jgens

You can't subtract it out like that. Remember that it's multiplied to e3. If you're not familiar with this, you should probably review your elementary algebra.

Anyway, before this goes further, the approach that you're taking (dividing through by x) doesn't really work. You should be able to get the equation 3x2+x-e3 = 0 and then solve for x.

7. Sep 10, 2010

### Ericalvusa91

I suppose I use the quadratic formula but I'm only suppose to get one answer

I get (sqrt (12e^3))/(6) and -(sqrt (12e^3))/(3)

8. Sep 10, 2010

### jgens

Think about which of those values make sense for log(x). The domain of the functions involved is very important.

9. Sep 10, 2010

### Staff: Mentor

Yes, use the quadratic formula. Since your first equation has ln(3x + 1) and ln(x), x must be positive, which will probably eliminate one of the solutions you get from the quadratic formula.

10. Sep 10, 2010

### Ericalvusa91

Nope the positive value is wrong too

11. Sep 10, 2010

### jgens

That would be because you applied the quadratic formula incorrectly.

12. Sep 10, 2010

### Ericalvusa91

Just to be sure c would be -e^3 right?

13. Sep 10, 2010

### jgens

That's correct.

14. Sep 10, 2010

### Ericalvusa91

Heres what i Have

3x^2+x-e^3=0

-b+-√((b^2) - 4ac)/2a

-1+-√(1^2)-4(3)(-e^3))/2(3)

-1+-√(1+12e^3)/6

Then two possibilities but only need the positive value

1st possibility = -1+1+√(12e^3)/6 = √(12e^3) / 6

15. Sep 10, 2010

### Staff: Mentor

You still have a mistake. If 3x^2 + x - e^3 = 0,
then
$$x = \frac{-1 \pm \sqrt{1 + 12e^3}}{6}$$

One of these values is positive and the other is negative.

16. Sep 10, 2010

### Ericalvusa91

The second value is (-1/3)+√(12e^3)/(6) so should I find the lcd

17. Sep 10, 2010

### Staff: Mentor

How are you getting that?

You have 1 + 12e^3 in the radical, and there's no way you can simplify that.

18. Sep 10, 2010

### Ericalvusa91

I thought you can take the radical one

-1-√(1+12e^3)/6

I still get it wrong though

19. Sep 10, 2010

### Staff: Mentor

The two solutions to the quadratic equation are
x1 = [-1 + √(1+12e^3)]/6
and x = [-1 - √(1+12e^3)]/6

The first one above is positive; the second one is negative.

I don't know what you mean by this:

20. Sep 10, 2010

### Ericalvusa91

I thought I could separate √(1) and √12e^3.

Those x-values do not give me the right answer unless I am suppose to do something after that. However, if I already solved for x it should be the answer though right?