More about $\log(Fib(n))$ element of $\theta(n)$, as seen in these previous posts...

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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