SOLUTION: log2(4x^2+8x+4)

Algebra.Com
Question 1149390: log2(4x^2+8x+4)
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.

Notice that  4x^2 + 8x + 4 = 4*(x^2 + 2x + 1) = 4*(x+1)^2


Therefore,  


     =  = use properties of logarithms =  +  = 2 + .    ANSWER



Notice that in the last term, my expression includes  absolute value  |x+1|  under the logarithm sign.


It is because, although  (x+1)^2  is always positive,  (x+1)  CAN BE NEGATIVE.


Therefore, to avoid a catastrophe under the logarithm sign, I introduced  |x+1|  in my last term.



It is NECESSARY (!),  and  if you do not make it, your answer will be WRONG (!)


Actually, it is the most important moment in the solution of this problem, and the entire problem 
is designed and intended for you would learn it from my post (!)

Solved, explained and completed.


RELATED QUESTIONS

8x^2-4x-4 (answered by jim_thompson5910)
Hello! I'm having trouble with this logaritm:... (answered by lwsshak3)
solve for x:... (answered by MathLover1,greenestamps,MathTherapy)
#1. Log2(3x-7)+log2(x+2)=log2(x+1) #2.... (answered by CPhill,ikleyn)
log2(x + 2) + log2(x - 4) =... (answered by lwsshak3)
Log2(3x-4)+log2(5x-2)=4 (answered by Fombitz)
-7=-9-log2(4x+4) I got stuck with the -log2.. don't know where to go after... (answered by ikleyn)
log2(x - 2) + 5 = 8 - log2... (answered by mukhopadhyay)
log2(x - 2) + 5 = 8 - log2... (answered by Nate)