Prove, without a calculator, that $2^{1+\sqrt{5}}\neq 9$ The original question was: Show that there does not exist a strictly increasing function f : N → N satisfying
f (2) = 3 and f (mn) = f (m)f (n) for all m, n ∈ N.
But the solution eventually came down to having to prove that $2^{1+\sqrt{5}}\neq 9$. It seems really easy, but I couldn't make any meaningful progress. Does anyone have a "nice" way to show that the above statement holds true?
 A: If you want to use a bazooka to kill a fly: the Gelfond-Schneider theorem shows that $2^{1+\sqrt{5}}$ is transcendental.
A: Since $\sqrt{5}>2.2=\frac{11}{5}$, $$2^{1+\sqrt{5}}>2^{1+11/5}=2^{16/5}$$
And $2^{16/5}$ is greater than $9$, because $2^{16}=1024\cdot64>64000$ while $9^5=729\cdot81<750\cdot80=60000$.
A: Notice


*

*$2^8 = 256 > 243 = 3^5 \implies 2^{32} = (2^8)^4 > (3^5)^4 = 3^{20} = 9^{10} \implies 2^{3.2} \ge 9$.

*$5^3 = 125 > 121 = 11^2 \implies 5\sqrt{5} > 11 \implies \sqrt{5} > 2.2$.


Combine these, we have 
$$2^{1+\sqrt{5}} > 2^{3.2} > 9$$
A: Let $\tau=(1+\sqrt{5})/2$.  Suppose $2^{\tau}=3$ which is equivalent to the claim in the question.  Taking a cue from the Fibonacci sequence, render $2^{8-5\tau}=256/243>1$.    Alas, $8/5<\tau$ (ratios of consecutive Fibonacci numbers are alternately below and above $\tau$, and $8/5$ is in a "below" position) which forces $2^{8-5\tau}<1$.
A: If you know that $\log_{10} 3 \approx 0.477,\log_{10} 2 \approx 0.30103$ and $\sqrt 5 \approx 2.236$ you can derive $\log_2 (3^2) \approx 2 \frac {0.477}{0.30103}\approx 3.17$ while $1+\sqrt 5 \gt 3.23$
