Question

write the expression 2/3 [ln(x) + ln(x^3-2) -ln(x+4)] as a logarithm of a single quantity.

Answer 1

lnx+lny=ln(xy)
lnx-lny=ln(x/y)
xlny=ln(y^x)

(2/3)(lnx+ln(x^3-2)-ln(x+4))=
(2/3)(ln((x)(x^3-2))-ln(x+4))=
(2/3)($\frac{ln((x)(x^3-2))}{ln(x+4)}$=
($\frac{2ln((x^4-2x))}{3ln(x+4)}$=
($\frac{ln((x^3-2x)^2)}{ln((x+4)^3)}$