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

1 Answer

Oct 27, 2016

#x=e-1#

Explanation:

In general #ln(a)=c# means #e^c=a#
therefore
#color(white)("XXX")ln(x+1)=1#
means
#color(white)("XXX")e^1=x+1#

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

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