How to prove that the sequence $\ln(F_n)$ isn't on a line? $F_n$ = Fibonacci sequence

How to prove that the sequence $\ln(F_n)$ isn't on a line?

I noticed that the Fibonacci sequence is not a geometrical sequence:
$$\frac{F_2}{F_1}=\frac{1}{1}=1\neq 2=\frac{2}{1}=\frac{F_3}{F_2}$$
Things I've done so far:
Lemma 1 The graph of the function $\ln(f(x))$ is a line $\iff f(x)=\mu \cdot t^{\lambda x} , \ \ \ \   \mu,\lambda, t\in \mathbb{R}$ 
Proof
($\implies$) Let $\ln(f(x))$ be a line $\implies \ln(f(x))=kx  +l \ \ /\exp(x)$
$\iff f(x)= \exp(kx+l) = \exp(l)\cdot \exp(kx)\implies t=e, \lambda =k, \mu=\ exp(l)$
( $\Longleftarrow$ ) Let $f(x)=\mu \cdot t^{\lambda x} \ \ /\ln()$ 
$\iff \ln(f(x))=\ln(\mu \cdot t^{\lambda x})=\ln(\mu)+\ln(t^{\lambda x})=\ln(\mu)+\lambda x\cdot \ln(t)=(\lambda\cdot\ln(t))x+\ln(\mu)=kx+l$ and that's the equation of a line.

Lemma 2 $\mu\cdot t^{\lambda n}$ is a geometrical sequence
Proof
Let $g_n = \mu\cdot t^{\lambda n}$ and $n\in\mathbb{N}$ be arbitrary. We have: 
$$\frac{g_n}{g_{n-1}}=\frac{\mu\cdot t^{\lambda n}}{\mu\cdot t^{\lambda (n-1)}}=\frac{t^{\lambda n}}{t^{\lambda (n-1)}}=\frac{(t^{\lambda})^n}{(t^{\lambda})^{n-1}}=(t^\lambda)^{n-(n-1)}=t^\lambda \ \  \forall n\in\mathbb{N}$$
$\implies g_n$ is a geometrical sequence with the corresponding quotient being $q=t^\lambda$

Lemma 3 If $a_n$ is not a geometrical sequence, neither is $a_n^\lambda \ , \ \lambda\in\mathbb{R} \setminus\{0\} $
Proof
Let us assume the opposite i.e. $a_n$ is not a geometrical sequence but $\exists \lambda\in\mathbb{R} \setminus\{0\}$ so that $a_n^\lambda$ is a geometrical sequence.
$a_n^\lambda$ is a geometrical sequence $\implies \forall n\in\mathbb{N} \ \frac{a_n^\lambda}{a_{n-1}^\lambda}=q\in\mathbb{R}$
$a_n$ is not a geometrical sequence $\implies \exists \  m_1,m_2\in\mathbb{N}$ so that $$\frac{a_{m_1}}{a_{m_1-1}}=q_1 \neq q_2=\frac{a_{m_2}}{a_{m_2-1}}$$
We have:
$$q=\frac{a_{m_1}^\lambda}{a_{m_1-1}^\lambda}=\left(\frac{a_{m_1}}{a_{m_1-1}}\right)^\lambda=q_1^\lambda$$ 
$$q=\frac{a_{m_2}^\lambda}{a_{m_2-1}^\lambda}=\left(\frac{a_{m_2}}{a_{m_2-1}}\right)^\lambda=q_2^\lambda$$ 
$$q_1^\lambda=q=q_2^\lambda \iff q_1^\lambda=q_2^\lambda \iff q_1=q_2$$ which is a contradiction so the assumption was false, therefore $a_n^\lambda$ is not a geometrical sequence.

The original "plan" was to notice that
$$\mu \cdot t^{\lambda x} = \mu \cdot (t^x)^\lambda \longrightarrow \sqrt[\lambda]{\mu \cdot (t^x)^\lambda} = \sqrt[\lambda]{\mu}\cdot t^x = \alpha\cdot t^x \longrightarrow \alpha\cdot t^n$$
and use the fact that the Fibonacci sequence is not geometrical so it can't be written as $\alpha\cdot t^n$ and by extension, it can't be written as $\mu\cdot t^{\lambda n}$ and therefore its range is not a subset of a function of the form $\mu\cdot t^{\lambda x}$ so (by Lemma 1) the sequence $\ln(F_n)$ is not on a line.
I thought I could use Lemma 3 to justify $F_n \longrightarrow \alpha\cdot t^n \longrightarrow \mu\cdot t^{\lambda n}$ but I think that I only managed to confuse myself even more. I'm not sure how to make the connection $F_n  \longrightarrow \mu\cdot t^{\lambda n}$. I'm not even sure if that's what I'm looking for. 
I tried messing with the idea of a direct connection to $f(x)$ i.e. $F_n  \longrightarrow \mu\cdot t^{\lambda x} \longrightarrow \ln(f(x))\longrightarrow p ... kx + l$ 
$n\longrightarrow F_n \longrightarrow \ln(F_n) \longrightarrow x_n$
$1\longrightarrow 1 \longrightarrow 0 \longrightarrow x_1$
$2\longrightarrow 1 \longrightarrow 0\longrightarrow x_2$
$3\longrightarrow 2 \longrightarrow \ln(2) \longrightarrow x_3$
$4\longrightarrow 3 \longrightarrow \ln(3) \longrightarrow x_4$  
$kx_1 + l =0 \ \ \ \ \ \ \ \ \ (1)$
$kx_3 + l =\ln(2)\ \  (2)$
$(2)-(1) : (x_3-x_1)k=\ln(2)$
but I didn't manage to notice anything useful.
Thanks in advance for your replies!
 A: The following proof works for all sequences of integers satisfying $F_{n+1}= F_n + F_{n-1}$. Recall that such sequences satisfy
$$\lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \frac{1 + \sqrt{5}}{2}$$
which is an irrational number.
Argue by contradiction, and suppose that all points $(n , \ln F_n)$ lie on a line. This means that there exist two constants $a,b$ such that $$\ln F_n = an+b$$
this is equivalent to
$$F_n = e^{an+b} = (e^a)^n \cdot e^b = A^n \cdot B$$
where we put $A=e^a$, $B=e^b$.
Now, consider ratios of the form $F_{n+1}/F_n$. By our condition, we have for all $n$
$$\frac{F_{n+1}}{F_n} = A$$
which is a rational number, and it is constant. This contradicts the irrationality of 
$$\frac{1 + \sqrt{5}}{2} = \lim_{n \to \infty} \frac{F_{n+1}}{F_n} = A$$
(note that the hypothesis may be relaxed to arguing that the points $(n, \ln F_n)$ eventually lie on a line).
A: We have$$(\forall n\in\mathbb{Z}^+):F_n=\frac{\varphi^n-(-1/\varphi)^n}{\sqrt5},\tag1$$where $\varphi$ is the golden ration, and therefore$$(\forall n\in\mathbb{N}):\log(F_n)\simeq n\log(\varphi)-\log\left(\sqrt 5\right).$$So, the only line that could possibly be an answer to your problem would be that line $y=\log(\varphi)x-\log\left(\sqrt 5\right)$. But this line won't work precisely because of the presence of $(-1/\varphi)^n$ in $(1)$.
