Can we choose integers \(x\) and \(y\) so that:
\[\log_6 x +\log_6 y =1?\]
How many different ways are there to do this?
If we use the statement \[\log_x ab = \log_x a + \log_x b\] we can rewrite the equation, \[\log_6 x+\log_6 y=1,\] as \[\log_6 (x \times y)= 1.\]
Knowing that \(\log_x x = 1\) we can see that in the equation above \(x \times y = 6\).
Six has two pairs of factors: \(1 \times 6\) and \(2 \times 3\), so there are two possible ways to choose integers \(x\) and \(y\) so that \(\log_6 x + \log_6 y = 1\):
\[\log_6 1 + \log_6 6 = 1\] and \[\log_6 2 + \log_6 3 = 1.\]
Why did we not consider pairs of integers \(x\) and \(y\) such as \(-1 \times -6\)?
In theory, if we thought about evaluating each term in the equation we might try to choose a value of \(x\) that is a power of \(6\). We might choose something like \(x=36\) so that \(\log_6 36 = 2\).
What is the problem that we would encounter if we tried this approach?
What further information does this give us about the values of \(x\) and \(y\)?
How about using \(\log_{12}\)?
Taking the same approach as above we have that \[\log_{12} x + \log_{12} y = \log_{12} (x \times y) = 1.\]
By the same reasoning as the previous example we can conclude that \(x \times y = 12\) and as \(12\) has three pairs of factors there will be three possible ways to choose integers \(x\) and \(y\) so that \(\log_{12} x + \log_{12} y = 1\):
\[\log_{12} 1 + \log_{12} 12 = 1,\]
or
\[\log_{12} 2 + \log_{12} 6 = 1,\]
or
\[\log_{12} 3 + \log_{12} 4 = 1.\]
How about using \(\log_{24}\)?
\(24\) has four pairs of factors so there will be four possible ways to choose integers \(x\) and \(y\) so that \(\log_{24} x + \log_{24} y = 1\):
\[\log_{24} 1 + \log_{24} 24 = 1,\] \[\log_{24} 2 + \log_{24} 12 = 1,\] \[\log_{24} 3 + \log_{24} 8 = 1,\]
and
\[\log_{24} 4 + \log_{24} 6 = 1.\]
In all of the examples above we have only considered positive values of \(x\) and \(y\). Why?
