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)(
`(ln((x)(x^3-2)))/(ln(x+4))`=
(
`(2ln((x^4-2x)))/(3ln(x+4))`=
(
`(ln((x^3-2x)^2))/(ln((x+4)^3))`