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

1 Answer

Mar 5, 2016

#=>x=e^4-1#

Explanation:

#lnsqrt(x+1)=2#
#=>log_esqrt(x+1)=2#
#=>sqrt(x+1)=e^2#
#=>(x+1)=e^4#
#=>x=e^4-1#

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