Help with solving an in-equation with a variable as exponent

Question

I need to decide on an integer value for $n$ in the following in-equation:

$\left(1-\dfrac{1}{2^{64}}\right)^n \leq \dfrac{1}{2}$

What I have tried:

The expression above reminds me of the known limit:

$\lim\limits_{n \to \infty}\left(1 - \dfrac{1}{n}\right)^n = \dfrac{1}{e} $

So, my intuition is to just determine $n = 2^{64}$ and to evaluate the entire left-hand-side expression as approaching to $\dfrac{1}{e}$ and since that is less than $\dfrac{1}{2}$ then it "feels" true

However, I feel this is really not a the way of solving this equation and there can probably be a better way to do it and obviously get a better value for $n$.

Thanks in advance

Answer 1

Since $\log$ function is increasing, we have that

$$\left(1-\frac{1}{2^{64}}\right)^n \leq \frac{1}{2} \iff n\log\left(1-\frac{1}{2^{64}}\right) \leq \log \frac{1}{2} \iff n\ge\frac{\log \frac{1}{2}}{\log\left(1-\frac{1}{2^{64}}\right)}\approx1.28\cdot 10^{19}$$

Answer 2

If OP wants some $n$ to satisfy the inequality they can proceed as follows. From the inequality $$ e^x\geq 1+x $$ for all $x$ we have that $$ \left(1-\frac{1}{2^{64}}\right)^n\leq e^{-n/2^{64}} $$ and $$ e^{-n/2^{64}}\leq \frac{1}{2} $$ if $$ n\geq 2^{64}\times \log 2\geq 2^{32} $$ Hence $n=2^{32}$ works for example.