SOLUTION: Solve Log6(x+3) + Log6(x-2) = 1 * 6 is the base for both logs
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Question 283616
:
Solve
Log6(x+3) + Log6(x-2) = 1
* 6 is the base for both logs
Answer by
user_dude2008(1862)
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Log6 (x+3) + Log6 (x-2) = 1
Log6 [(x+3)(x-2)] = 1
(x+3)(x-2) = 6^1
(x+3)(x-2) = 6
x^2+x-6 = 6
x^2+x-12=0
(x+4)(x-3)=0
x+4=0 or x-3=0
x=-4 or x=3
x=-4 is extraneous
Answer: Only solution is x=3