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 $1$, $\log_21010_2\approx3.25$
Best Answer
It depends on what you have memorized. For $\log_2 10$, many people know that $\log_{10} 2\approx 0.30103$, so $\log_2 10 = \frac 1{0.30103}=\frac {10}{3}-\frac 13\% \approx 3.32$ where the last $\frac 13\%$ is optional and accounts for the difference between $0.3$ and $0.301$. The correct value is about $0.3322$
Many base $2$ logs are easier because you might know some powers of $2$
For $\log_3 10$ we know $3^2=9$ so it should be a little more than $2$. I think my next step would be to say $3^{2.5}=9*1.732=17.32-1.73=15.59$ As $10$ is about $\frac 17$ of the way from $9$ to $16$ I would add $\frac {0.5}7=0.07$ and say $\log_3 10 \approx 2.07$ The correct value is about $2.10$ Not too bad for a long linear interpolation.
I did play fair and not check the correct result until afterward. These are fairly hard as estimates go. Multiplication and division are much easier as one is more prone to have useful facts memorized. $\ln 10=2.30$ and $\ln 2=0.693$ can be useful, too.