For any positive number \(x\), can you explain why this result is true?

\[2\log_4 x=\log_2 x\]

Can you describe what \(\log_4 x\) and \(\log_2 x\) are in words?

\(\log_4 x\) is the power that we’d need to raise \(4\) to if we wanted to get \(x\), and \(\log_2 x\) is the power that we’d need to raise \(2\) to if we wanted to get \(x\).

Which of the following is correct?

\[2\log_6 x=\log_3 x \quad 3\log_9 x=\log_3 x \quad 2\log_9 x=\log_3 x\]

Thinking about how logarithms are defined, the equation \(2\log_4 x=\log_2 x\) seems to hold because \(4\) is \(2^2\), rather than because \(4\) is \(2\) ‘doubled’. Therefore of the three alternatives, \(2\log_9 x=\log_3 x\) is likely to be the equivalent result, but for base \(3\) and \(9\). To check this, let \(\log_{9} x = k.\) Then \(x=9^{k}\), so \[\log_{3}x= \log_{3}9^{k}=\log_{3}3^{2k}=2k= 2\log_9 x.\]

What is a general statement of this result?

There are various relationships here that we can start to generalise.

How is \(\log_{a}x\) related to \(\log_{a^{2}}x\) ?

How is \(\log_{a}x\) related to \(\log_{a^{n}}x\)?

How is \(\log_{a}x\) related to \(\log_{b}x\)? How can we express the relationship between \(a\) and \(b\)?

From the examples above, we can see that \[ \log_{a}x= 2\log_{a^{2}}x,\] which we could generalise to \[\log_{a}x= n\log_{a^{n}}x.\] Thinking about \(n\)’s role in this equation brings us to

\[\log_a x=(\log_a b)(\log_b x)\] or equivalently \[\log_{b}x=\frac{\log_{a}x}{\log_{a}b}.\]

What conditions must \(x,a,b\) satisfy here? Could the second of these ever risk division by \(0\)?