Between 2 and 3.
\log_10(270) log10(270)
= \log_10(2.7 \times 10^2)=log10(2.7×102)
= \log_10 (2.7) + \log_10(10^2)=log10(2.7)+log10(102)
=\log_10(2.7) + 2=log10(2.7)+2
Now, \log(2.7)log(2.7) lies somewhere between 0 and 1 because 2.7 lies somewhere between 10^0100 and 10^1101. In mathematical terms, 10^0 < 2.7 < 10^1 \iff \log(10^0) < \log(2.7) < \log(10^1)100<2.7<101⇔log(100)<log(2.7)<log(101). Therefore 2 + \log(2.7)2+log(2.7) must lie between 2 + 02+0 and 2 + 12+1 which is 22 and 33.
Similarly, you can directly state it this way:
10^2 < 270 < 10^3 102<270<103
\iff \log(10^2) < \log(270) < \log(10^3)⇔log(102)<log(270)<log(103)
\iff 2 < \log(270) < 3⇔2<log(270)<3
The latter seems to work better. I just introduced the former because it forces a student to actually think of it in terms of the orders of magnitude.
