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 \(\longrightarrow\) | |||||
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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.
