Log Fibonacci = Theta

Question

I'm trying to prove that $\log F_n = Θ(n)$ and where

$$F_n = F_{n-1} + F_{n-2}$$ $F_1 = 1$, $F_0 = 0$

There's already a thread about this question, but the accepted answer doesn't explain a certain calculation in detail, and I was hoping someone here understands and could explain it to me!

I don't understand the part

$$\log F_{n+1}=\log(F_n+F_{n-1})=\log F_n+\log\left(1+{F_{n-1}\over F_n}\right)$$

More precisely I don't understand how $log(F_{n-1}) = \log\left(1+{F_{n-1}\over F_n}\right)$

I hope someone can explain how this is possible.

Answer 1

Since $F_{n+1}=F_n+F_{n-1}$, and $\ln (x\cdot y)=\ln x+\ln y$, $$\ln F_{n+1}=\ln(F_n+F_{n-1})=\ln(F_n\cdot(1+\frac{F_{n-1}}{F_n}))=\ln F_n+\ln(1+\frac{F_{n-1}}{F_n}).$$

Answer 2

\begin{align}\ln(F_n+F_{n-1})=\ln F_n+\ln\left(1+{F_{n-1}\over F_n}\right)&\impliedby \small\exp(\ln(F_n+F_{n-1}))=\exp(\ln F_n)\cdot\exp\left(\ln\left(1+{F_{n-1}\over F_n}\right)\right)\\&\impliedby F_n+F_{n-1}=F_n\cdot\left(1+{F_{n-1}\over F_n}\right)\\&\impliedby F_n+F_{n-1}=F_n+\require{cancel}\cancel{F_n}\frac{F_{n-1}}{\cancel{F_n}}\end{align}