SOLUTION: Hi :) I�m new to logarithms and not sure if I�m on the right track with this one. Hoping you can help?... thanks heaps log₃x + log₃(x + 2) = 1 (find x) My attempt:

Question 255027: Hi :)
I�m new to logarithms and not sure if I�m on the right track with this one. Hoping you can help?... thanks heaps
log₃x + log₃(x + 2) = 1 (find x)
My attempt:
> log₃((x )(x+2)) = 1
> log₃(x�+2x) = 1
> x� + 2x = 3�
> x� + 2x � 3 = 0
> (x - 1)(x + 3) = 0
> so x = +1 & -3
Is that right???
thanks again
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log₃x + log₃(x + 2) = 1 (find x)
My attempt:
> log₃((x )(x+2)) = 1
> log₃(x�+2x) = 1
> x� + 2x = 3�
> x� + 2x � 3 = 0
> (x - 1)(x + 3) = 0
> so x = +1 & -3
.
So far, so good!
At this point, you must "test" for "extraneous" solutions.
Notice that if x was equal to -3
log₃(x + 2) = log₃(-3 + 2) = log₃(-1)
You cannot take a log of a negative number --
therefore, -3 is an extraneous solution -- throw it out leaving:
x = 1
as your solution
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SOLUTION: Hi :) I�m new to logarithms and not sure if I�m on the right track with this one. Hoping you can help?... thanks heaps log&#8323;x + log&#8323;(x + 2) = 1 (find x) My attempt:

SOLUTION: Hi :) I�m new to logarithms and not sure if I�m on the right track with this one. Hoping you can help?... thanks heaps log₃x + log₃(x + 2) = 1 (find x) My attempt: