Find least integral value of n such that $$\left(\frac{\root \of 5+1}{2}\right)^n > 100$$
I want to solve this without the use of a calculator or logarithmic tables. I tried using AM-GM, such that the "lesser than" side is equal to $100$. $$\left(\frac{\root \of 5+1}{2}\right)^n > 5^{\frac{n}{4}}$$ Now $5^\frac{n}{4} = 100$. Again, same issue, I am unable to solve this without log tables.
On
The interesting thing here is you have the golden ratio, $\newcommand{\p}{\varphi} \p$. It satisfies the identity $$ \p^2 = \p+1 $$ and hence $$\begin{align*} \p^3 &= \p^2 + \p = 2\p+1 \\ \p^4 &= 2\p^2 + \p = 2(\p+1)+\p = 3\p+2 \\ \p^5 &= 3\p^2 + 2\p = 3(\p+1)+2\p = 5\p + 3 \end{align*}$$ and so on. One could naturally extrapolate this to a pattern of $$ \p^n = F_n \p + F_{n-1} $$ for $F_n$ the $n$th Fibonacci number.
It is then easy to proceed by exhaustion, really. One sees that $$ \p^9 = 34 \p + 21 \qquad \p^{10} = 55\p+34 $$ One could use a crude approximation of $\p$, say $\p \approx 1.618$, to comfortably conclude that the $n$ desired is $10$.
Here's a slightly entertaining way to do it - perhaps not what's intended!
Binet's formula says that the $n$th Fibonnaci number is given by
$$F_n = \frac{1}{\sqrt{5}}(\varphi^n - \psi^n)$$ where
$\varphi = \frac{1}{2}(1 + \sqrt{5})$ and $\psi = \frac{1-\sqrt{5}}{2}$. Since $|\psi| < 1$, we can approximate $F_n \approx \frac{1}{\sqrt{5}} \varphi^n$, with an error $< 1$.
So we're reduced to finding the least $n$ such that $F_n > 100/\sqrt{5}$, which is about $45$, again to the nearest integer. This is then an easy task - $n = 10$.