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To demonstrate that this result below is true, but for all logarithm base greater than 1, can we just say that 'Logarithms in different bases differ only by a constant multiplier' ?

$$\frac{n}{2}\log(2) = n\log(\sqrt{2}) \leq \log(F_n) \leq n\log(2).$$

Thanks

Previous posts: Show that log $Fib_{n}$ is $\theta(n)$ Log Fibonacci Equals to Theta Log Fibonacci = Theta

Nick
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Emanual
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  • What you want to prove, $\log F_n=\theta(n)$, is that $\log F_n \in [an,bn]$. It follows from $F_n=F_{n-1}+F_{n-2},F_n\ge F_{n-1}$, $F_n\le 2F_{n-1}\le 2^n F_1, F_n\ge 2 F_{n-2}\ge 2^{n/2-1}F_1$. – reuns Feb 11 '20 at 02:58
  • In https://math.stackexchange.com/questions/2971350/show-that-log-fib-n-is-thetan?noredirect=1&lq=1, Cain told something that I think is not true at all... $$F_n \geq (\sqrt{2})^n$$ As I can see: F(0)=0 is not greater nor equal than $$(\sqrt{2})^0 = 1$$ F(1)=1 is not greater or equal than $$(\sqrt{2})^1 = (\sqrt{2})$$ F(2)=1 is not greater nor equal than $$(\sqrt{2})^2 = 2$$ ... – Emanual Feb 12 '20 at 01:45
  • @reuns: I don't really understand the last part: $$ F_n\le 2F_{n-1}\le 2^n F_1, F_n\ge 2 F_{n-2}\ge 2^{n/2-1}F_1$$ – Emanual Feb 12 '20 at 01:49

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