Summing to one

Summing to one

Excerpt: This task explores the behaviour of logarithms to different bases, by asking for solutions to equations in which sums of logarithms are equal to one.

Problem 1

$$\log_3 3=1$$ $$\log_9 3+ \log_9 3 =1$$ $$\log_{27} 3 + \dots +\log_{27} 3= 1$$

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

Solution 1

$$\log_3 3=1$$ $$\log_9 3+ \log_9 3 =1$$ $$\log_{27} 3 + \dots +\log_{27} 3= 1$$

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 $4\log_{81}3= 1.$

Problem 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?

How about using $\log_{12}$?

How about using $\log_{24}$?

How might your answers change if the equations were instead of the form $\log_n x - \log_n y = 1?$

Solution 2

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?