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

1 Answer
iceman
Sep 14, 2015

#x=e^2-2#

Explanation:

#lnsqrt(x+2)=1=># If: #log_a(x)=yhArrx=a^y#, then:

#sqrt(x+2)=e^1 = e=>#square both sides:

#x+2=e^2=>#subtract 2 from both sides:

#x=e^2-2=>#

All roots must be checked in the original equation to verify that they work and are not "Extraneous" roots that were introduced during the squaring process, so:

#lnsqrt(x+2)=1=># at: #x=e^2-2#
#=lnsqrt(e^2 - 2 + 2)#
#=lnsqrt(e^2)#
#=ln(e)#
#=1#
#1=1=>#hence verified: #x=e^2-2#, is a valid root.

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