SOLUTION: I am trying to solve logarithmic and exponential functions. However, I have gotten stuck. What is the solution to log(3x+7) + log(x-2)=1

SOLUTION: I am trying to solve logarithmic and exponential functions. However, I have gotten stuck. What is the solution to log(3x+7) + log(x-2)=1

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Question 591874: I am trying to solve logarithmic and exponential functions. However, I have gotten stuck. What is the solution to log(3x+7) + log(x-2)=1
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
adding logs means multiplying their arguments; and one is the log of 10

(3x + 7)(x - 2) = 10

remember that logs are not defined for negative arguments
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