Question #984b4

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Answer 1 · 2017-04-14T07:31:31

#x=sqrt(1/(e-1))#

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

By Log Property: #rln x=ln x^r#,

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

By raising #e# to both sides,

#Rightarrow e^(ln(1+x^2))=e^(1+ln x^2)=e^1 cdot e^(ln x^2)#

By Inverse Property: #e^(ln x)=x#,

#Rightarrow 1+x^2=ex^2#

By subtracting #x^2# from both sides,

#Rightarrow 1=ex^2-x^2=(e-1)x^2#

By dividing both sides by #(e-1)#,

#Rightarrow 1/(e-1)=x^2#

By taking the square-root of both sides,

#pm sqrt(1/(e-1))=x#

Since the domain of #ln x# is #x>0#, we have

#x=sqrt(1/(e-1))#

I hope that this was clear.