'How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?
I'm still wondering about this; if anyone gets close.. just tell me and I'll tweak your answer more. Thanks in advance!
Hello, Masta X!
Is this close enough?How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?
We have: .$\displaystyle \frac{2^{21}}{3\cdot10^5} \;=\;\frac{2,\!097,\!152}{300,\!000} \;=\;
6.990506667 $
Therefore: .$\displaystyle 21\log2 - \log 3 - 5 \;=\;0.844508654 \;\approx\;\log 7$