How do you solve $ln x - ln (x + 1) = 1$ $lnx-ln(x+1)=1$ ?

Question

3816 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-23T07:03:11

no solution

Since $ln a - ln b = ln (\frac{a}{b}) and ln e = 1$ $ln a - ln b=ln(a/b) and lne=1$

you have the equivalent equation:

$ln (\frac{x}{x + 1}) = ln e$ $ln(x/(x+1))=lne$

Then:

$\frac{x}{x + 1} = e$ $x/(x+1)=e$

You also must fix:

$x > 0 and x + 1 > 0$ $x>0 and x+1>0$

that is $x > 0$ $x>0$

and

$x = e (x + 1)$ $x=e(x+1)$

$x = e x + e$ $x=ex+e$

$x - e x = e$ $x-ex=e$

$x (1 - e) = e$ $x(1-e)=e$

$x = \frac{e}{1 - e}$ $x=e/(1-e)$

but it is a negative number, against the domain of the solution fixed as a positive number, then the equation has no solution

How do you solve lnx−ln(x+1)=1 lnx-ln(x+1)=1?