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
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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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scott8148(6628)
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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