# Math Help - Logistic Problem - Help Please!

1. ## Logistic Problem - Help Please!

'How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?

2. Originally Posted by .Masta X
'How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?

Hint: Remember that
$\log(xy)=\log(x)+\log(y)$

3. Hello, Masta X!

How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?
Is this close enough?

We have: . $\frac{2^{21}}{3\cdot10^5} \;=\;\frac{2,\!097,\!152}{300,\!000} \;=\;
6.990506667$

Therefore: . $21\log2 - \log 3 - 5 \;=\;0.844508654 \;\approx\;\log 7$

4. Yes, Soroban!
I got something like that, but I was off by 0.0020.

Thank you very much! I really appreciate your help greatly!