How do you use the properties of logarithms to expand #ln((x^2-1)/x^3)#?
1 Answer
Nov 19, 2017
Explanation:
If
#ln(a/b) = ln(a) - ln(b)#
and:
#ln(ab) = ln(a)+ln(b)#
Hence we also find:
#ln(a^n) = ln(overbrace(a * a * ... * a)^"n times") = overbrace(ln(a) + ln(a) + ... + ln(a))^"n terms" = n ln(a)#
So:
#ln((x^2-1)/x^3) = ln(x^2-1)-ln(x^3)#
#color(white)(ln((x^2-1)/x^3)) = ln((x-1)(x+1))-3ln(x)#
#color(white)(ln((x^2-1)/x^3)) = ln(x-1) + ln(x+1) - 3ln(x)#
