Logarithm lineup

Initial observations

• We can write down exact values for two of these logarithms.
• Since $\log_3 5$ is the power of $3$ that would give us $5$ and $3^1<5<3^2$ we know the value of $\log_3 5$ must lie between $1$ and $2$. Can you do the same for other logarithms in the list and use the results to put some of them in order?

Logarithms in ascending order so far

Increasing or decreasing?

If we take a series of logarithms all with the same base, we might assume that $\log$ of a bigger number will have a bigger value. In other words, the function $\log x$ is increasing.

If it is true for the logarithms in our list, we can use this fact to put some of them in order.

Changing bases

If instead, we take a series of logarithms of the same number but to different bases, would these increase or decrease? Think about $\log_2 8$ and $\log_4 8$. Can we use this idea to order some of our logarithms?

Another way to compare these logarithms of the same number would be to use the rule for changing the base: $$\log_a b=\log_a c \times \log_c b\text{ .}$$

For instance, to compare $\log_3 5$ and $\log_4 5$, we could write $\log_3 5 = \log_3 4\times\log_4 5$ and then note that $\log_3 4 > 1$, and so $\log_3 5 > \log_4 5$.

Ascending order diagram

Our network of orderings now looks like this.

Estimating values

It is possible to find some other upper and lower bounds within our list of logarithms by using approximations for the square roots of $2$ and $3$:

We know that $2^\frac{1}{2}\approx1.4$, so $2^\frac{3}{2}=2\times2^\frac{1}{2}\approx2.8$.

Therefore $\frac{3}{2}\approx\log_2{2.8}$, and hence $\log_2{3}>\frac{3}{2}$.

Can you use an approximate value of $\sqrt 3$ to find an upper bound for $\log_3 5$?

Increasing or decreasing again

We shall now look at two possible ways of comparing $\log_3 4$ and $\log_4 5$ which is the last unknown pairing in our lineup.

Firstly, notice that in each of these remaining logarithms, the base is one less than the number whose logarithm we are taking. We could think of them as values of the function $\log_x{(x+1)}$. Is this an increasing or decreasing function? Think about its value when $x=2$ and when $x$ is very large. You could use Desmos to explore it.

Solution

This appears to be a decreasing function for $x>1$ which is the domain we are interested in. When $x$ gets very large, the value of this function will be very close to $\log_x {(x)}=1$ and when $x=2$ its value is $\log_2 3$ which is certainly bigger than $1$.

Different differences

Another way to compare $\log_4 5$ and $\log_3 4$ would be to work out the difference between them, $$d=\log_3 4 - \log_4 5 \text{ .}$$ If we can show that the difference is positive then we can put the two logarithms in order.

Can you use the subtraction law of logarithms to rearrange this and show that $d>0$?

Solution

We can complete the argument like this: $$\begin{align*} d&=(\log_3 4 - \log_3 3)-(\log_4 5 -\log_4 4)\\ &=\log_3 {\frac{4}{3}}-\log_4 {\frac{5}{4}} \end{align*}$$ Now, $\log_4 {\frac{5}{4}} < \log_3 {\frac{5}{4}}$ because $\log_x {\frac{5}{4}}$ is a decreasing function for $x>1$. Therefore, $$\begin{align*} d&>\log_3 {\frac{4}{3}}-\log_3 {\frac{5}{4}}\\ &=\log_3 {\Big(\frac{\big(\frac{4}{3}\big)}{\big(\frac{5}{4}\big)}\Big)}\\ &=\log_3 {\big(\frac{16}{15}\big)}\\ &>0 \end{align*}$$

So now we have shown that $d$ is positive, so we can put the two original logarithms in order.

Sometimes, adding zero can help you solve a problem! You may wonder how we knew that subtracting one from each side would help. The answer is that we didn’t know, but…

Looking at this numberline fragment, we already knew the two logarithms were between $1$ and $2$. By subtracting the ones, we have effectively compared the two red dotted lengths rather than the two blue dashed lengths. What is it that makes that feel like a worthwhile thing to do?

Final solution

Finally we have a complete lineup of logarithms.