# Natural Log Equation

## Homework Statement

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

## 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?

HallsofIvy
Homework Helper
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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Ok, thanks! :D

I just wanted to be sure I hadn't missed out on anything.

gb7nash
Homework Helper
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.

Ok, got it, thanks! :D

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.

HallsofIvy
Homework Helper
I don't believe that was ever in question!

Delta2
Homework Helper
Gold Member
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.

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.

Mentallic
Homework Helper
If you punch $$10^{99}-(10^{99}-1)$$ 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.

HallsofIvy