Mental Math - Estimating Logarithms 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 $1$, $\log_21010_2\approx3.25$

What I do is consider instead $\log_2 10^3$ (for example), if I wish to calculate $\log_2 10$. The number $2$ has the nice property that $2^{10}$ is "roughly" $1000$, which makes $$\log_2 10 = \frac{1}{3}\log_2 10^3 \simeq \frac{1}{3}\log_2 2^{10} = \frac{10}{3} \simeq 3.3.$$ I also have the unhealthy habit of calculating powers of $2$ from time to time in my head, $1024,2048,4096,8192...$
For powers of $3$, the same strategy holds $3^{21} \simeq 10^{10}$, which gives us $\log_3 10\simeq 2.1$. The whole point of these arguments is that one can get rid of the decimal parts for a while (for mental calculation purposes).
