How do you solve #ln (x+1)^2 = 2#?

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Answer 1 · 2018-06-21T06:12:32

We know that,

#color(red)(lnX=Y=>log_eX=Y <=> X=e^Yto(1)#

Here,

#ln(x+1)^2=2#

#=>(x+1)^2=e^2...tocolor(red)(Apply(1)#

#=>x+1=e#

#=>x=e-1#

Answer 2 · 2018-06-21T09:15:50

#color(crimson)(x = e - 1 = 1.7183#

#ln(x+1)^2 = 2#

#cancel2 * ln(x+1) = cancel2#

#ln(x+1) = 1#

#x + 1 = e^1 = e#

#x = e - 1#