I have this simple question I saw here:
±(2 - 2^(-23)) × 2^128 = ±6.8 × 10^38
How did they get to ±6.8 × 10^38 ?
I tried :
(2 - 2^(-23)) × 2^128 $\Longrightarrow$ (2^1 - 2^(-23)) × 2^128 $\Longrightarrow$ 2^129 - 2^(105) $\Longrightarrow$ now what ?
On
Something very useful is to notice that $2^{10} = 1024 \approx 10^3$ You calculations are almost correct (wrong sign in front of 105) but you can neglect $2^{105}$ because it is very small compared to $2^{129}$ So :$$2^{129} = 2^9(2^{10})^{12}\approx 512\times (10^3)^{12} = 5.12\cdot 10^2\times10^{36}$$
This way you can find a very good estimate of you answer with light calculations.
If you really want to find the $6.8$ instead of $5.12$, then use the fact that $(1+x)^n\approx 1+nx$ for small $x$ ie :$$1.024^{12} \approx 1 + 0.024*12 = 1.288$$
$1.024^{12}$ is what I neglected in my above approximation so if you plug that in the result I had above you obtain: $$2^{129} \approx 1.288 \times 5.12\times10^{38} \approx 6.6\times 10^{38}$$
By taking a better approximation of $1.024^{12}$ you can achieve the wanted result.
What is nice with this method is that there is no need for a calculator, all the calculations can be done by hand easily.
Let's ignore the insignificant part $2^{105}$ first
$$ x = 2^{129} \\ \log_{10} x = 129 \times \log_{10}2 \approx 38.83286944 \\ \therefore x \approx 10^{38.83286944} = 10^{0.83286944} \times 10^{38} \approx 6.805647 \times 10^{38} $$
Similarly, we can get $2^{105} \approx 4.056 \times 10^{31}$.