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:

Algebra ->  Exponential-and-logarithmic-functions -> 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:       Log On


   

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(x²+2x) = 1
> x² + 2x = 3¹
> x² + 2x – 3 = 0
> (x - 1)(x + 3) = 0
> so x = +1 & -3
Is that right???
thanks again
Answer by nerdybill(7384) About Me  (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(x²+2x) = 1
> x² + 2x = 3¹
> x² + 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