-lg (1/u)
don't understand this with the - log and how they come to the result lg(u) hope you can help me
\(\text{suppose }v=\log(u) \\ \text{then }e^v = u \\ e^{-v} =\dfrac{1}{e^v} = \dfrac 1 u \\ -v = \log\left(\dfrac{1}{u}\right)\)
