How Do I Simplify ln(4+4y+4y^2)?

[m0286]

hi quick question for this:
simplify the following

ln(4+4y+4y^2) i know i need to use those simplifying rules for ln which are ln(xy)=lnx+lny
ln(1/x)=-lnx
ln(x/y)=lnx-lny
ln(x^r)=rlnx

but for the question I am asked to simplify.. it doesn't really follow any of these... I was wondering what the rule is for this type of question. Do i do:
ln4+ln4y+ln4y^2?? Thanks!

 

[ircdan]

Notice 4 + 4y + 4y^2 = 4(1 + y + y^2), now use the fact that ln(ab) = lna + lnb.

but for the question I am asked to simplify.. it doesn't really follow any of these... I was wondering what the rule is for this type of question. Do i do:
ln4+ln4y+ln4y^2?? Thanks!

Careful here! It is not true in general that ln(a + b) = lna + lnb. Goodluck.

 

[m0286]

ooops wrote the question wrong!

woopsies wrote the question wrong... its actually ln(4+4y+y^2)
so I arranged the inside so (y+2)(y+2) so can I use ln(ab) = lna + lnb
so that ln(y+2)(y+2)=ln(y+2)+ln(y+2)? Would that we right? Thanks

 

[arildno]

Certainly!
And, furthermore, this is consistent with:
\log((y+2)^{2})=2\log(y+2)

 

[VietDao29]

Certainly!
And, furthermore, this is consistent with:
\log((y+2)^{2})=2\log(y+2)

There should be an absolute value there. :)
It should read:
\ln ((y + 2) ^ 2) = 2 \ln |y + 2|
Note that the domain of the LHS is all the reals but y = -2, so the domain of the RHS must also be all the reals but y = -2. And if you are working in the reals, ln(y + 2) is only defined for y > -2. :)

 

[arildno]