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-lg (1/u)

don't understand this with the - log and how they come to the result lg(u) hope you can help me

 Oct 11, 2018
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\(\text{suppose }v=\log(u) \\ \text{then }e^v = u \\ e^{-v} =\dfrac{1}{e^v} = \dfrac 1 u \\ -v = \log\left(\dfrac{1}{u}\right)\)

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 Oct 11, 2018

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