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

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Answer 1 · 2015-12-20T13:19:51

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

Given: $ln (x + 1) - ln (x) = 1$ $ln(x+1)-ln(x)=1$

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

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

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

$x + 1 = x \cdot e$ $x+1 = x*e$

$x - x \cdot e = - 1$ $x-x*e = -1$

$x \cdot (1 - e) = - 1$ $x*(1-e)=-1$

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

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