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

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Answer 1 · 2015-12-20T13:19:51

$x=1/(e-1)$

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

$ln((x+1)/x)=1$

$e^(ln((x+1)/x))=e^1$

$(x+1)/x=e$

$x+1 = x*e$

$x-x*e = -1$

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

$x=1/(e-1)$

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