Come risolvi #ln (x + 1) = 1 #?

Generalmente #ln(a)=c# si intende #e^c=a#
perciò
#color(white)("XXX")ln(x+1)=1#
si intende
#color(white)("XXX")e^1=x+1#

#color(white)("XXX")rarr x=e-1#