Solve x: Calculus Natural Log Homework Statement

[Ericalvusa91]

Homework Statement

ln(3x+1) = 3-lnx

Homework Equations

Solve for x

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 ?

 

[jgens]

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

So I divide the x

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

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.

 

[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

 

[jgens]

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

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

 

[Ericalvusa91]

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

I subtracted the x^-1

 

[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.

 

[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)

 

[jgens]

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

 

[Mark44]

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.

 

[Ericalvusa91]

Nope the positive value is wrong too

 

[jgens]

That would be because you applied the quadratic formula incorrectly.

 

[Ericalvusa91]

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

 

[jgens]

That's correct.

 

[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

 

[Mark44]

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.

 

[Ericalvusa91]

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

 

[Mark44]

The second value is (-1/3)+√(12e^3)/(6)

How are you getting that?

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

 

[Ericalvusa91]

How are you getting that?

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

I thought you can take the radical one

-1-√(1+12e^3)/6

I still get it wrong though

 

[Mark44]

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:

I thought you can take the radical one

 

[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?

 

[Mark44]

Are you thinking that √(a + b) = √a + √b? That is not true.

Why do you think that those values are incorrect? (Only one works in your original equation, though.)

The positive value should make your original equation a true statement. If so, then you're done.

 

[Ericalvusa91]

Thank You. I finally got it! Instead of putting the -1 divided by 6, I left it as whole number 1.