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Logx/log(x-8) = log6x
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\[ \log(A) + \log(B) = \log(A*B) \\ \text{ } \\ \text{If log(C) = log(D), then C = D} \\ \]
\[ \log(x) + \log(x-2) = \log(3x) \\ \log(~x * (x-2)~) = \log(3x) \\ x * (x-2) = 3x \\ x^2 - 2x - 3x = 0 \\ x^2 - 5x = 0 \\ x(x-5) = 0 \\ x = 0 \text{ or } x = 5 \\ \text{Try each solution in the original equation. } \\ \text{x = 0 will not work because log(0) is undefined. } \\ \text{x = 5 will work.} \\ x = 5 \]
ohhhh ok I see now. Thank you
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