# What numbers would the base-10 logarithm of 270 fall between?

Oct 7, 2014

Between 2 and 3.

$\setminus {\log}_{10} \left(270\right)$
$= \setminus {\log}_{10} \left(2.7 \setminus \times {10}^{2}\right)$
$= \setminus {\log}_{10} \left(2.7\right) + \setminus {\log}_{10} \left({10}^{2}\right)$
$= \setminus {\log}_{10} \left(2.7\right) + 2$

Now, $\setminus \log \left(2.7\right)$ lies somewhere between 0 and 1 because 2.7 lies somewhere between ${10}^{0}$ and ${10}^{1}$. In mathematical terms, ${10}^{0} < 2.7 < {10}^{1} \setminus \iff \setminus \log \left({10}^{0}\right) < \setminus \log \left(2.7\right) < \setminus \log \left({10}^{1}\right)$. Therefore $2 + \setminus \log \left(2.7\right)$ must lie between $2 + 0$ and $2 + 1$ which is $2$ and $3$.

Similarly, you can directly state it this way:

${10}^{2} < 270 < {10}^{3}$
$\setminus \iff \setminus \log \left({10}^{2}\right) < \setminus \log \left(270\right) < \setminus \log \left({10}^{3}\right)$
$\setminus \iff 2 < \setminus \log \left(270\right) < 3$

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.