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Natural Log Equation

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Homework Statement



Given the Equation ln(x+1)-ln(x)=0 find x (If it exists.

Homework Equations





The Attempt at a Solution



ln(x+1)-ln(x)=0

ln(x+1)=ln(x)

x+1 = x

1 = x-x

Solution does not exist. (My book "agrees" with me in the solutions.)

However, typing up the equation in my TI-84 and using the equation solver it finds a solution at around 9.785*10^98. (Which, in fact, works.)

Am I doing something wrong?
 

Answers and Replies

  • #2
HallsofIvy
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No, it doesn't work. For x= 9.875 x 10^90, ln(x+1)- ln(x) is about 0.0907, not 0. I don't know exactly what you did but I suspect that the "10^98" caused a round off error. I did ln(10.785 x 10^98)- ln(9.785 x 10^98) the "hard way" first, ln(10.785 x 10^98)= ln(10.785)+ 98/ln(10),and the same for ln(9.785 x 10^98, and then simply as ln(10.785 x 10^98/9.785 x 10^98)= ln(10.785/9.785)= 0.0907..., not 0. There is no real x satifying ln(x+1)- ln(x)=0.
 
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  • #3
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Ok, thanks! :D

I just wanted to be sure I hadn't missed out on anything.
 
  • #4
gb7nash
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However, typing up the equation in my TI-84 and using the equation solver it finds a solution at around 9.785*10^98. (Which, in fact, works.)
If you're wondering why you were getting 0, it was roundoff error. Depending on what computer system you use, subtracting two values that are relatively the same will give you 0.
 
  • #5
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Ok, got it, thanks! :D
 
  • #6
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No, it doesn't work. For x= 9.875 x 10^90, ln(x+1)- ln(x) is about 0.0907, not 0. I don't know exactly what you did but I suspect that the "10^98" caused a round off error. I did ln(10.785 x 10^98)- ln(9.785 x 10^98) the "hard way" first, ln(10.785 x 10^98)= ln(10.785)+ 98/ln(10),and the same for ln(9.785 x 10^98, and then simply as ln(10.785 x 10^98/9.785 x 10^98)= ln(10.785/9.785)= 0.0907..., not 0. There is no real x satifying ln(x+1)- ln(x)=0.
just to clarify 9.785 x 10^98 + 1 is NOT equal to 10.785 x 10^98.
 
  • #7
HallsofIvy
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I don't believe that was ever in question!
 
  • #8
Delta2
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Yes there is a "kind of" solution x=infinite... your calculator just approximate infinite by a very large number of the order 10^98 ... But ofcourse infinite cannot be considered as a solution to an equation.
 
  • #9
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I don't believe that was ever in question!
It definitely wasn't. But in your calculations of ln(x+1) - ln(x), u have calculated ln(10.785/9.785). Though it doesn't change the logic behind your explanation, I was just clarifying that the numbers are wrong.
 
  • #10
Mentallic
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If you punch [tex]10^{99}-(10^{99}-1)[/tex] into your calculator you will get 0. Again just another demonstration of roundoff errors.

This is another reason why it's sometimes a good idea to do some easy simplifications to an expression before punching it in.
 
  • #11
HallsofIvy
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It definitely wasn't. But in your calculations of ln(x+1) - ln(x), u have calculated ln(10.785/9.785). Though it doesn't change the logic behind your explanation, I was just clarifying that the numbers are wrong.
Finally, the coin drops! Of course you are right. Thanks.
 

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