SOLUTION: Please help me with the following problem:
2log(y+2)=log(y+2)-log12
What I tried -
log(y+2)-log12-log(y+2)^2
y+2/12(y+2)^2=1
y+2/12y^2+48y+48=1
y+2=12y^+48y+48
12y^2+47y
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Please help me with the following problem:
2log(y+2)=log(y+2)-log12
What I tried -
log(y+2)-log12-log(y+2)^2
y+2/12(y+2)^2=1
y+2/12y^2+48y+48=1
y+2=12y^+48y+48
12y^2+47y
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Question 60825: Please help me with the following problem:
2log(y+2)=log(y+2)-log12
What I tried -
log(y+2)-log12-log(y+2)^2
y+2/12(y+2)^2=1
y+2/12y^2+48y+48=1
y+2=12y^+48y+48
12y^2+47y+46=0
That's where I got! - please help! Answer by Nate(3500) (Show Source):
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