How do you condense $ln(x^(3) - 8) - ln(x^(2) + 2x + 4) + ln(x + 2)$ ?

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Answer 1 · 2018-07-09T18:39:19

$ln(x^2-4)$

We get

$ln(x^3-8)+ln(x+2)=ln((x^3-8)(x+2))$
after the law: $ln(a)+ln(b)=ln(ab)$ for $a,b>0$

then we get

$ln(((x^3-8)(x+2))/(x^2+2x+4))$
now we use that

$a^3-b^3=(a-b)(a^2+ab+b^2)$
so we get

$x^3-8=(x-2)(x^2+2x+4)$
after cancelling like Terms we get

$ln(x^2-4)$

How do you condense ln(x^(3) - 8) - ln(x^(2) + 2x + 4) + ln(x + 2)ln(x^(3) - 8) - ln(x^(2) + 2x + 4) + ln(x + 2)?