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 ,

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

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

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

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

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

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

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

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

$=>x^2=0$

$x=0$

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