$\log_2(2^{64} - 2^{32})$
I have tried to do $64 - 32 = 32$ but $\log_2{32} = 5?$
Thx for help in forward advance
Edit: Apologies for my poor spelling and grammar. Please forgive me. I am new to this forum and I joined in because I want to be smart so I can get 6 in gcse maths.
$\log_2(2^{64}-2^{32})$ is very close to $64$ -- so close that the difference may not show on the calculator you're using. With a high-precision calculator I get
$$ \log_2(2^{64}-2^{32}) = 63.999\,999\,999\,664\,096\,384\,7\ldots $$
We can predict this without calculating in detail, because $$ 2^{64}-2^{32} = 2^{64}(1-2^{-32}) $$ and therefore $$ \log_2(2^{64}-2^{32}) = 64 + \log_2(1-2^{-32}) $$ Here $1-2^{-32}$ is very close to $1$, so its logarithm is very close to $0$ and hardly makes a dent in the $64$.
For a next approximation we can write $$ \log_2(2^{64}-2^{32}) = 64 + \log_2(1-2^{-32}) = 64 + \frac{\ln(1-2^{-32})}{\ln 2} $$ and use the approximation $\ln(1+x)\approx x$ when $x$ is small. Here we have $x=-2^{-32}$ which is certainly small, so $$ \log_2(2^{64}-2^{32}) \approx 64 - \frac{2^{-32}}{\ln 2} \approx 63.999\,999\,999\,664\,096\,385\,0 $$ which is extremely close to the result above.