Summing to one

Add to your resource collection

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 $$\log_6 2 + \log_6 3 = 1.$$

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

$$\log_{12} 3 + \log_{12} 4 = 1.$$

$24$ 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

$$\log_{24} 4 + \log_{24} 6 = 1.$$