Question 973910: The integral of log3 (x^2) - log3 (x+6)= 1. After solving this I find that x= -3 and x = 6, which is correct. My question is whether or not -3 is a solution?
when testing the original question should it be:
A) log3 (-3^2) - log3 (-3 + 6) =1 in which case -3 is a solution, or should it be
B) 2log3 (-3) - log3 (-3 +6) =1, in which case -3 is not a solution because the first term, log3 (-3), does not exist? Thanks
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! You have the correct solution.
(log(x^2))/(log(3))-(log(x+6))/(log(3)) = 1. Now I multiply both sides by log(3)
log(x^2)-log(x+6) = log(3)etcetera. You've already done the problem so it saves me some writing.
You factor the left and get:
(x-6) (x+3)= 0
x-6= 0 or x+3= 0
x= 6 or x= -3 This is your solution
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