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Unusual log problem

1. Homework Statement

Solve for "x" in the equation: ln(x+2)-ln(x+3)+ln(x)=1



3. The Attempt at a Solution

This unusual ln problem really has me stumped with what to do and usually this stuff has given me no problems. I can't isolate the x because e is always in the way of solving it. From "ex+3e=x^2+2x" the answer I got was "-2+2(1+3e)^1/2/2e" (?)
 

Answers and Replies

454
0
e is just a number, so ex+3e=x^2+2x is just a quadratic. put it in the form ax^2 + bx +c and apply the quadratic formula.
 
I've already done that which simplifies to the answer shown in the first post. I set it equal to ex and then divided by e. I'm not clear as to why it equalled to zero at the end which can't possibly make any physical sense

Another approach I used was to treat 2x-ex as "b" although that produced:

-2-e+(e^2+4+8e)^1/2/2e

That still makes it wrong (ugh)
 
Last edited:
454
0
The answer in your first post is wrong. you'll have to give more details to tell what went wrong. You just want to collect terms with x^2, x and things indendent of x, and then to apply the quadratic formula.
 
335
0
Take ln x to R.HS. and express 1 as 2.303*log10/2.303

simplify.
 
The quad formula doesn't seem to work in any of the cases and I'm extremely reluctant to change e to its actual irrational form because it'll just get messier.

Physixguru, I'm scratching my head here - where did you get 1= 2.303*log10/2.303?

Kamerling, for the first trial I thought about just isolating the x. So, ex=x^2+2x-3e. Then I tried x^2+2x-ex-3e where b is 2x-ex
 
Last edited:
Gib Z
Homework Helper
3,344
4
Take kammerlings original suggestion again. Treat e like any other constant, pretend you know nothing about its numerical value. You should get [itex]x^2 + (2-e) x - 3e = 0[/itex], a simple quadratic equation.
 
335
0
ln x= 2.303 log x (assuming base 10 )

now 1= 2.303/2.303 ( log 10)
 
Here is how I did it this time- all the actual steps

-2-e+( (2-e)^2-4(1)(-3e) )^1/2/2

-2-e+( e^2-4e+4+12e )^1/2/2

-2-e+( e^2+8e+4 )^1/2/2

-2-e+( 4(1+2e) )^1/2/2

-2-e+2e(1+2e)^1/2/2

-2+e(1+2e)^1/2/2

Can someone confirm if this is right?
 
Last edited:
HallsofIvy
Science Advisor
Homework Helper
41,738
898
Take ln x to R.HS. and express 1 as 2.303*log10/2.303

simplify.
ln x= 2.303 log x (assuming base 10 )

now 1= 2.303/2.303 ( log 10)
None of this makes any sense at all. Apparently "physixguru" is advocating changing to common logs. I have no idea why.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
898
Here is how I did it this time- all the actual steps

-2-e+( (2-e)^2-4(1)(-3e) )^1/2/2
[-(2-e)+/- ( (2-e)^2-4(1)(-3e) )^1/2]/2= [-2+e+/- ( (2-e)^2-4(1)(-3e) )^1/2]/2
better:
[tex]\frac{-(2-e)\pm\sqrt{(2-e)^2- 4(1)(-3e)}}{2}= \frac{-2+e\pm\sqrt{e^2+ 8e+ 4}}{2}[/tex]
Notice the sign difference.

-2-e+( e^2-4e+4+12e )^1/2/2

-2-e+( e^2+8e+4 )^1/2/2

-2-e+( 4(1+2e) )^1/2/2

-2-e+2e(1+2e)^1/2/2

-2+e(1+2e)^1/2/2

Can someone confirm if this is right?
 
Last edited by a moderator:
Thanks HallsofIvy! :-) Made my day as usual
 
16
0
would you like an accurate approach to this problem, ArtofScience?
 
16
0
x = 3.2373049

The answer is quite simple to obtain. You had it right when you stated xˆ2 + 2x = ex + 3e.

Now just set the equation = 0, so xˆ2 + (2-e)x - 3e = 0...

At this point, it's only a matter of graphing it on your calculator or plugging it into the quad form. Either way, you will still obtain the answer of 3.2373049. To check this answer, just plug it back into the original equation for x and be amazed at how it works!
 
HallsofIvy
Science Advisor
Homework Helper
41,738
898
In what sense is that more accurate than the solution ArtofScience gave?
 

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