Solving Unusual Log Problem: Find x in ln(x+2)-ln(x+3)+ln(x)=1

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In summary, the conversation discusses how to solve for "x" in the equation ln(x+2)-ln(x+3)+ln(x)=1. Multiple approaches are suggested, including treating e as a constant and using the quadratic formula. Ultimately, the conversation concludes that the solution is x = 3.2373049, which can be confirmed by plugging it back into the original equation.
  • #1
The_ArtofScience
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Homework Statement



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



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" (?)
 
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  • #2
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.
 
  • #3
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)
 
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  • #4
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.
 
  • #5
Take ln x to R.HS. and express 1 as 2.303*log10/2.303

simplify.
 
  • #6
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
 
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  • #7
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.
 
  • #8
ln x= 2.303 log x (assuming base 10 )

now 1= 2.303/2.303 ( log 10)
 
  • #9
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?
 
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  • #10
physixguru said:
Take ln x to R.HS. and express 1 as 2.303*log10/2.303

simplify.

physixguru said:
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.
 
  • #11
The_ArtofScience said:
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?
 
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  • #12
Thanks HallsofIvy! :-) Made my day as usual
 
  • #13
would you like an accurate approach to this problem, ArtofScience?
 
  • #14
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!
 
  • #15
In what sense is that more accurate than the solution ArtofScience gave?
 

1. How do I approach solving this log problem?

To solve this problem, you will need to use the properties of logarithms. Specifically, you will need to use the product, quotient, and power rules of logarithms to manipulate the equation and isolate the variable x.

2. Can I solve this problem by hand or do I need a calculator?

This problem can be solved by hand using the properties of logarithms and algebraic manipulation. However, if you are not comfortable with solving logarithmic equations, you may find it helpful to use a calculator.

3. What is the importance of the natural logarithm (ln) in this problem?

The natural logarithm (ln) is used in this problem because the equation is written in terms of natural logarithms. The ln function is the inverse of the exponential function e^x, and it is commonly used in mathematical and scientific applications.

4. Are there any restrictions on the values of x in this problem?

Yes, there are restrictions on the values of x in this problem. Since the logarithm of a negative number is undefined, the expression inside the logarithms (x+2, x+3, x) must be positive. This means that x cannot be less than -2 or -3, otherwise the equation would be undefined.

5. Can I check my solution to this problem?

Yes, you can check your solution to this problem by plugging in your answer for x into the original equation. If the equation is true, then your solution is correct. You can also use a graphing calculator to plot the original equation and your solution to visually verify the answer.

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