Building blocks

## Solution 2

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?