math question, plz help?

Question

ln a = 2, ln b = 3, ln c = 5

find,

ln ( a^-1 b^-4)/ (ln (bc)^3)

thanks for your help

Answer 1

ln(a^-1 b^-4)=lna^-1+lnb^-4 (because lnxy=lnx+lny)
=-lna-4lnb=-2-12=-14 (because lnx^z=z lnx)
ln(bc)^3=3lnbc=3lnb+3lnc=9+15=24

ln(a^-1 b^-4)/(ln(bc)^3)=-14/24=-7/12

Answer 2

To answer this you need to know these laws of lns:

1: ln(xy) = lnx + lny
2: ln(a^x) = xlna

I'll do the first bit for you...


ln ( a^-1 b^-4)/ (ln (bc)^3) =
Using Law 1:

(ln(a^-1) + ln(b^-4))/ ln(b^3c^3)=

(ln(a^-1) + ln(b^-4))/ (ln(b^3) + ln(c^3))


Now you need to use law 2 to simplify it further ie
ln(a^-1) = -1lna

Answer 3

ln ( a^-1 b^-4)/ (ln (bc)^3)
= [ln(1/a)+ln(1/b^4)][3ln b+ 3lnc]
[ln1-lna +ln1-4lnb][3ln b+ 3lnc]
=[0 -2 +0 -4*3][3*3+3*5]
= -14/24 = -7/12