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I have the following equation:

$\log(S_n) = \log(u)[2T-n]\,\,$

I was just wondering how $S_n = u^{2T-n}$ is then obtained?

Thank You

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By taking the exponential of both sides. Recall that $e^{a\ln b}=b^a$.

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You can also use the fact that $a \times \log{x} = \log{(x^a)}$ which means that $$\log{(u)} [2T-n] = [2T-n] \times \log{(u)} \\ = \log{(u^{[2T-n]})} = \log {S_n} \Rightarrow S_n = u^{[2T-n]}$$

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