SOLUTION: Solve for x: log (2 lower)(x+2) + log (2lower)x=3 log (2 lower)[x(x+2)] = 3 2^3=x(x+2) 8=x^2 + 2x x^2 + 2x-8=0 (x+4)(x-2)=0 x=-4, x=2 log(2 lower)(-4 + 2) + log (2 lower)

Question 73211: Solve for x: log (2 lower)(x+2) + log (2lower)x=3
log (2 lower)[x(x+2)] = 3
2^3=x(x+2)
8=x^2 + 2x
x^2 + 2x-8=0
(x+4)(x-2)=0
x=-4, x=2
log(2 lower)(-4 + 2) + log (2 lower) (-4)=3
this is where i get stuck and need some help. Thanks!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log (2 lower)[x(x+2)] = 3
2^3=x(x+2)
8=x^2 + 2x
x^2 + 2x-8=0
(x+4)(x-2)=0
x=-4, x=2
log(2 lower)(-4 + 2) + log (2 lower) (-4)=3
this is where i get stuck and need some help.
------------
You found that x=-4 or x=2
You began to check the x=-4 answer and got stuck.
Right.
x=-4 cannot be an answer of the original problem because log(-4) has no meaning.
--------
So check x=2 to see if it is a good answer.
log(2+2) + log2 = 3
Using base 2 you get:
2+1=3
Which is true.
So x=2 is a good answer and is the only answer.
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Cheers,
Stan H.
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