How can we estimate logarithms with **different bases**? Take $\log_2 10$ ($1\over\log_{10}2$$\approx3.32192809$) for example. If we convert $10$ to binary, we get $1010_2$. So $\log_21010_2$ can clearly be estimated at between $3$ and $4$ because $\log_21000_2 = 3$ because $1000_2 = 8_{10}$.

How can we estimate the decimal part?

**Edit:** The decimal part can be estimated by seeing how close it is to the integer part. See following examples.

More examples:

$\log_3 10_{10} = \log_3101_3 \approx 2 $

$\log_4 10_{10} = \log_422_4 \approx 1.5 $

$\log_510_{10}=\log_520_5\approx1.5$

$\log_610_{10}=\log_614_6\approx1.25$

$\log_{11}10_{10}=\log_{11}A_{11}\approx1$

$\log_{11}100_{10}=\log_{11}91_{11}\approx1.9$

Calculating $\log_21010_2$:

$1010 \to 101.0 \to 10.10 \to 1.010$. Because we did it 3 times, $\log_21010_2\approx3$. Because $.01\approx1/4$ of the way to $1$, $\log_21010_2\approx3.25$