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

2 Answers
Jun 21, 2018

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

Jun 21, 2018

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

Explanation:

ln(x+1)^2 = 2

cancel2 * ln(x+1) = cancel2

ln(x+1) = 1

x + 1 = e^1 = e

x = e - 1