Natural Log Question - Solve for x...

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rufusbabe

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Here is the question:

Solve for x:
ln(x+1) - lnx + lnx2 = 1

My working so far (not sure if it's right...)

ln(x+1) - lnx + 2lnx = 1

ln(x+1) + lnx = 1

ln(x(x+1) = 1

ln(x2 + x) = 1

That's where I've got to... I was thinking I have to use the quadratic equation to solve for x but do I need to get rid of the ln first and make the equation equal 0?

Any help with this would be awesome. Thanks in advance. :)
 
rufusbabe said:
Here is the question:

Solve for x:
ln(x+1) - lnx + lnx2 = 1

My working so far (not sure if it's right...)

ln(x+1) - lnx + 2lnx = 1

ln(x+1) + lnx = 1

ln(x(x+1)) = 1

ln(x2 + x) = 1

That's where I've got to... I was thinking I have to use the quadratic equation to solve for x but do I need to get rid of the ln first and make the equation equal 0?

Any help with this would be awesome. Thanks in advance. :)
Click to expand...
So far so good - except it would have been clearer if you used parentheses to show that lnx^2 meant ln(x^2).

Exponentiate both sides of your final step. That is, the inverse operation of "ln". THEN you will have a quadratic in x. You will have to use the quadratic formula, because one of the terms will be irrational.
 
Thank DrPhil for the hint... but I think I'm stuck again... :(

ln(x2 + x) = 1

x2 + x = e1

x2 + x - 2.718281828 = 0

(the remaining is in the image as I'm not sure how to type it on here)

Assign2.jpg

So I end up with a square root of a negative number which is not possible... :(
What have I done wrong?
 
rufusbabe said:
Thank DrPhil for the hint... but I think I'm stuck again... :(

ln(x2 + x) = 1

x2 + x = e1

x2 + x - 2.718281828 = 0

(the remaining is in the image as I'm not sure how to type it on here)

View attachment 3513

So I end up with a square root of a negative number which is not possible... :(
What have I done wrong?
Click to expand...

For the quadratic equation, you have A = 1 and C = -2.718.. → AC = - 2.718 → b2 - 4AC = 1 + 10.872 = 11.872...
 
ummm, I haven't learnt that before....

I googled it and got - Extraneous Solution: A solution of a simplified version of an equation that does not satisfy the original equation.

Could you please explain which one is 'extraneous' and why for me?
 
Do I substitute the x values back in to the original equation to work out which one is 'extraneous'?
 
rufusbabe said:
ln(x2 + x) = 1
x2 + x = e1
x2 + x - 2.718281828 = 0
Click to expand...
Do no do that!

Solve x2+x−e=0x=−1±1+4e2\displaystyle \\x^2+x-e=0\\\\x=\dfrac{-1\pm\sqrt{1+4e}}{2}

e≠2.718281828\displaystyle e \ne {\rm{2}}.{\rm{718281828}}

 
Last edited:
Ok, so I get an error when I substitute the -2.22287023 value back into the original equation. So do I just discount this solution and leave the only solution as x = 1.22287023?
 
rufusbabe said:
Ok, so I get an error when I substitute the -2.22287023 value back into the original equation. So do I just discount this solution and leave the only solution as x = 1.22287023?
Click to expand...

Again do not use a decimal value for e\displaystyle e.

Also ln⁡(x)\displaystyle \ln(x) is defined only if x>0\displaystyle x>0.
 
pka said:
Do no do that!

Solve x2+x−e=0x=−1±1+e22\displaystyle \\x^2+x-e=0\\\\x=\dfrac{-1\pm\sqrt{1+e^2}}{2}

e≠2.718281828\displaystyle e \ne {\rm{2}}.{\rm{718281828}}
Click to expand...

Umm, I'm confused... if I do e1 on my calculator I get 2.718281828....

Also, How did you get e2​ in the quadratic equation you posted?
 
Ok, I edited my working to change the decimal back to e but I get the same numbers anyway....

Assign2.jpg

Is this wrong?
 
rufusbabe said:
I do e1 on my calculator I get 2.718281828....

Also, How did you get e2​ in the quadratic equation you posted?
Click to expand...

Your calculator is lying to you.
I know almost nothing about calculators.
Since the almost universal access to the internet at our university, I will not allow students to use them.

Look at this

There should be a button ex\displaystyle \boxed{e^x} and enter x=2\displaystyle x=2.
 
Thanks pka but I'm not trying to calculate e2... I'm calculating e1 since my original equation = 1

C for my quadratic equation equals -e1

See my full working...
Assign2.jpg

Unless I have made a mistake somewhere....
 
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