Solving Natural Logarithm Equation ln(x^2 + 1 ) -3lnx=ln2

[physstudent1]

Homework Statement

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

Solve for x.

Homework Equations

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.

 

[D H]

I used laws of logarithms to simplify it down to -x^2(x-1)=1 .

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

 

[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

 

[D H]

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

OK so far. You're mistake is here:

then I did

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

Can you see what you did wrong or do you need a bit more help?

 

[physstudent1]

Should I have made it

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

before doing the e^ ?

 

[D H]

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

 

[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

 

[D H]

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.

 

[Hurkyl]

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?

 

[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?

 

[Hurkyl]

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?

 

[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)

 

[Saladsamurai]

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)

Plug it in and see what happens.

 

[Hurkyl]

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)

 

[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