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Can anyone help me to understand this?

$$\log_2 n = \log_2e \log n$$

Daniel R
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Sam
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  • Change of base: $\log_r a = \frac{\log_s a}{\log_s r}$ and implicitly, in your notation: $\log = \log_e$, sometimes written as $\ln$. – StackTD Apr 20 '16 at 12:40
  • $$2^{\log_2n}=2^{\log_2e\log n}=(2^{\log_2e})^{\log n}=e^{\log n}.$$ –  Apr 20 '16 at 13:04
  • Do you want to understand the formula itself, or gain a feel for why logarithms in different bases are similar this way? – lauir Apr 20 '16 at 13:09

1 Answers1

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$\log_r(n)$ is the number such that $r^{\log_r(n)} = n$. Taking the $\log_s$ on both sides we have that $$ \log_s\left(r^{\log_r(n)}\right) = \log_s(n) \implies \log_r(n) \log_s(r) = \log_s(n). $$ In your case, let $s = 2$ and $ r = e $.

Lionel Ricci
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