We have extended one possible solution grid. Can you use other logarithms to complete the four blank cells of this grid or one based on your own solution grid?

Increasing size \(\longrightarrow\) | |||||

Increasing size \(\downarrow\) |
\(\log_{\square} \square\) | ||||

\(\log_{4} 2\) | \(\log_4 3\) | \(\log_{4} 5\) | |||

\(\log_{\square} \square\) | \(\log_{5} 3\) | \(1\) | \(\log_{3} 5\) | \(\log_{\square} \square\) | |

\(\log_{5} 4\) | \(\log_3 4\) | \(\log_{2} 4\) | |||

\(\log_{\square} \square\) |

Some things to think about

- Try to make your extra logarithms as general as possible. For instance if you want \(\log_{a} 3\) to be less than \(\log_{5} 3\) what conditions must \(a\) satisfy?
- How could you include fractions or irrational numbers in your logarithms?

In order to fill in the box at the left of the grid, we require a logarithm that is smaller than \(\log_{5} 3\). Two possible solutions are \(\log_{6} 3\) and \(\log_{5} 2\). Both of these are less than \(\log_{5} 3\).

In order to give a more general logarithm for this cell, let’s think about when \(\log_{5} a < \log_{5} 3\). We know \(a\) cannot be zero or negative because then it would not be possible to calculate its logarithm. In addition to this, \(\log_{5} a\) will only be smaller than \(\log_{5} 3\) when \(a < 3\). So we need \(0 < a < 3\).

Notice that \(a\) does not have to be a whole number—it could be a fraction or an irrational number. Note too that the value of the logarithm can be negative.

Alternatively, still thinking about the same cell, we could try a logarithm of the form \(\log_{b} 3\). This would be smaller than \(\log_{5} 3\) if \(b > 5\). Is it possible for the base of the logarithm, \(b\), to be a fraction or an irrational number? If so, are there any other values of \(b\) that would make \(\log_{b} 3\) smaller than \(\log_{5} 3\)?

Now suppose we wrote \(\log_{b} 4\) in the box at the bottom of the grid. This must be greater than \(\log_{3} 4\) so we need \(b < 3\), but this time there are other constraints on \(b\) because you cannot use \(1\), \(0\) or any negative number as the base of a logarithm. Also, if the base, \(b\) is between \(0\) and \(1\), then \(\log_b 4\) will be negative, so the correct constraint would be \(1 < b < 3\).

To understand fractional bases of logarithms, it might help to think about \(\log_{\frac{1}{2}} 4\). \[\big(\frac{1}{2}\big)^{-2}=(2)^2=4 \quad\text{ so }\quad \log_{\frac{1}{2}} 4=-2\]

Can you sketch the graph of \(y=\log_x 4\)? Try using graphing software such as Desmos (find \(\log_a\) on its functions menu). You might also find the resource Changing bases helpful for thinking about the behaviour of logarithmic functions with different bases.