How do you solve #Ln(e^3 - x^3) = ln(e^2 - x^2) +1#?

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Answer 1 · 2018-08-04T10:58:48

#x=0#

We know that

#(1)lnM-lnN=ln(M/N)#

#(2)lne=1#

Here ,

#ln(e^3-x^3)=ln(e^2-x^2)+1#

#=>ln(e^3-x^3)-ln(e^2-x^2)=1#

#ln((e^3-x^3)/(e^2-x^2))=lne#

#=>(e^3-x^3)/(e^2-x^2)=etoApply(1) and (2)#

#=>((e-x)(e^2+ex+x^2))/((e-x)(e+x))=e#

#=>((e^2+ex+x^2))/((e+x))=e#

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

#=>e^2+ex+x^2-e^2-ex=0#

#=>x^2=0#

#x=0#