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

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Answer 1 · 2018-06-16T12:03:13

THe solution is #=e^2-2#

Solve the equation as follows :

#lnsqrt(x+2)=1#

#=>#, #ln(x+2)^(1/2)=1#

#=>#, #1/2ln(x+2)=1#

#=>#, #ln(x+2)=2#

#=>#, #x+2=e^2#

#=>#, #x=e^2-2=5.389#

Answer 2 · 2018-06-16T12:13:52

#x=e^2-2~~5.3890560989#

Remember that #ln(a)=c# means #log_e(a)=c#
#color(white)("XXXXXXXXXXXXXXXXXX") rArre^c=a#

So
#color(white)("XXX")ln(sqrt(x+2))=1#
means
#color(white)("XXX")e^1=sqrt(x+2)#

and therefore
#color(white)("XXX")e^2=x+2#
and
#color(white)("XXX")x+2(color(magenta)(-2))=e^2color(magenta)(-2)#
or
#color(white)("XXX")x=e^2-2#

...an approximation to this value can be determined using a calculator.