\[\log_3 3=1\]

\[\log_9 3+ \log_9 3 =1\]

\[\log_{27} 3 + \dots +\log_{27} 3= 1\]

Perhaps some logical first thoughts would be that the second equation, \[\log_9 3+\log_9 3=1,\] can be rewritten as \[2\log_9 3 = 1,\] and so \[\log_9 3 = \frac{1}{2}.\]

We now have that \(\log_3 3=1\) and that \(\log_9 3=\frac{1}{2}\) but what is \(\log_{27}3\)?

Remember that we also know that \(\log_3 3=1\) because \(3^1=3\), and that \(\log_9 3=\frac{1}{2}\) because \(9^{\frac{1}{2}}=3\).

This is because of one of the defining statements about logarithms: \(\log_a x = y\) means \(x=a^y\) for real numbers \(x>0\) and \(y\).

We can now ask the question, to what power do we raise \(27\) to get the result \(3\)? If we think about powers of \(3\) then we can quickly confirm that \(27 = 3^3\) and so \(27^{\frac{1}{3}}=3\).

Now we have that \(\log_{27} 3 = \frac{1}{3}\) and therefore \[\log_{27} 3 + \log_{27} 3 +\log_{27} 3= 1,\] or equivalently, \[\log_{27} 3 + 2\log_{27} 3 = 1,\] or \[\log_{27} 3 + \log_{27} 9 = 1.\]

An alternative approach would be to use the statement \[\log_x ab = \log_x a + \log_x b.\] We can use this to rewrite the second equation, \[\log_9 3+\log_9 3=1,\] as \[\log_9 (3 \times 3) = \log_9 9 = 1.\]

Knowing that \[\log_x x = 1\] we can now think of the third equation as

\[\log_{27} (3a)=1\] where \(3a = 27\) and therefore \(a=9\) as before, \[\log_{27} 3 + \log_{27} 9 = 1.\]

How many \(\log_{81} 3\) do you need to add together to make one?

We can use exactly the same thinking to address this problem. We could consider \(81^n=3\) and recognise that \(81 = 3^4\) so \(n=\frac{1}{4}\).

This tells us that we will need four lots of \(\log_{81} 3\) to make one, \[4\log_{81}3=1.\]

This thinking relies on us recognising, or quickly deducing that \(81\) is the fourth power of \(3\). If the base of the logarithm were something less familiar, for instance \(2187\) (the seventh power of \(3\)), then this method is less convenient.

Alternatively we could have used the laws of logs to conclude that \[\log_{81} 3+\log_{81} b = \log_{81} (3b) = 1\] and so \(3b = 81\) and hence \(b = 27\).

Then we have \[\log_{81} 3 + \log_{81} 27 = 1.\] But, remember that the question asks how many lots of \(\log_{81} 3\) we need so we need to make one. This means that we need to write \(\log_{81} 27\) in terms of \(\log_{81}3\).

To do this we should first recognise that \(27 = 3^3\) so we can write that \[\log_{81}27 = \log_{81}3^3 = 3\log_{81}3.\]

We have used the statement \[\log_a x^y = y\log_a x.\]

Now we have \[\log_{81}3 + \log_{81}27 = \log_{81}3 + 3\log_{81}3,\] and therefore \[4\log_{81}3= 1.\]