SOLUTION: Solve Log6(x+3) + Log6(x-2) = 1 * 6 is the base for both logs

Question 283616: Solve
Log6(x+3) + Log6(x-2) = 1
* 6 is the base for both logs
Answer by user_dude2008(1862) (Show Source): You can put this solution on YOUR website!
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
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