About decimals and exponents

Question

I was reading about the speed of convergence of iterative methods. Then I came across the following.

How is it inferred here that "At least 126 terms are needed to ensure this accuracy for the linearly convergent sequence"?

Answer 1

The linearly converging sequence is $(0.5)^n$. To solve $$ (0.5)^n = 10^{-38} \text{,} $$ take logarithms to the base $2$ on both sides $$ -n = -38 \log_2 10 = -126.233\dots \text{.} $$ So, $(0.5)^n$ is a bit larger than $10^{-38}$ for $n = 126$ and then a bit smaller for $n = 127$.

Answer 2

For fun, a bit of arithmetic:

Note: $2^{10} =1024 > 10^3$, and $2^7 =128 > 10^2$.

Hence :

$2^{120} \gt 10^{36}$, and

$2^{120}×2^7 > 10^{36}×10^{2}$,

$2^{127} > 10^{38}$, or

$10^{-38} > 2^{-127}=(0.5)^{127}.$