Solve for x if ln (x + 1) = 1 + ln x.

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

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

$\ln (x + 1) = \ln ex$
 

Hence,
$x + 1 = ex$

$1 = ex - x$

$(e - 1)x = 1$

$x = \dfrac{1}{e - 1}$   ←   answer