How do you solve $ln [(\frac{x + 1}{x})] = 1$ $ln [((x+1) / x)]=1$ ?

Question

1690 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-02T11:51:39

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

We have:

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

Therefore, by the definition of the logarithm, we have:

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

Which we now solve for $x$ $x$ :

$x + 1 = e x$ $x+1 = ex$

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

Giving the solution:

$x = 1/(e-1)$

How do you solve ln [((x+1) / x)]=1ln[(x+1x)]=1ln [((x+1) / x)]=1?