SOLUTION: How do you solve this log? {{{log (6, w+3) + log (6, w - 2) = 2}}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How do you solve this log? {{{log (6, w+3) + log (6, w - 2) = 2}}}      Log On


   

Question 929298: How do you solve this log?
log+%286%2C+w%2B3%29+%2B+log+%286%2C+w+-+2%29+=+2
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log+%286%2C+%28w%2B3%29%29+%2B+log+%286%2C+%28w-2%29%29+=+2
log+%286%2C+%28w%2B3%29%28w-2%29%29+=+2
%28w%2B3%29%28w-2%29+=36
w%5E2%2Bw-6=36
w%5E2%2Bw-42=0
%28w-6%29%28w%2B7%29=0
Two solutions:
w-6=0
w=6
and
w%2B7=0
w=-7
However since this leads to negative arguments for the log functions, this is not a valid solution.
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.
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w=6

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
log+%286%2C+w%2B3%29+%2B+log+%286%2C+w+-+2%29+=+2+
log%286%2C%28w%2B3%29%28w-2%29%29= 2
(w+3)(w-2) = 36
w^2 + w -6 = 36
w^2 + w - 42 = 0
(w+7)(w-6)= 0 (w = -7 an extraneous root...results in undefined logarithm)
w = 6