Now it seems to be quite clear. We can see data for both small and large number of engines. What is more we observe an interesting linear-like relationship between the number of engines and the noise level. To see what we have actually done here, let us introduce a logarithm \(\log\): \[\log_b(a) = c \iff b^c = a,\]

for \(a > 0, b > 0\) and \(b\neq 1.\) For now let's not dwelve into meaning of all this. Let's instead remember the logarithm symbol \(\log\) and maybe also notice the fact, that we can treat it as a function, i.e. when you provide two numbers \(a\) and \(b\), logarithm spits out another number \(c\). For now we are also not much interested in the number \(b\). We may even omit it for some time and instead of writing \(\log_b(a),\) write simply \(\log(a)\).

Having introduced logarithm, we can say that in our case, the noise level \(l\) is a logarithm of number of engines \(n\), i.e: \[l = \log(n).\]

Let us see some properties of \(\log\), based on our example. First, remember that our reference value was when number of engines running was equal to \(1\). Since the increase in noise level in any subsequent cases will be referred to this reference value, we can safely assume it is equal to \(0\), i.e \(\log(1) = 0.\) Now, when the second engine was turned on, the noise level increased by \(\Delta l\). Therefore, \(\log(2) = \Delta l\). As you may remember, with \(4\) engines running, the noise level has again increased by \(\Delta l\), thus \(\log(4) = 2\Delta l\). The same goes for 8 engines, where we have to add \(\Delta l\) to the previous value \(\log(4)\) and obtain \(\log(8) = 3\Delta l\), etc. Now, you may notice that we have picked our engines turned in such a way that their number doubles. Therefore, each number \(2\), \(4\), \(8\)... is a power o \(2\). We have \(2 = 2^1\), \(4 = 2^2\), \(8 = 2^3\), \(16 = 2^4\) and so on. Combining these two observations, we get:

\begin{align}\log(2) &= \Delta l\\ \log(4) &= \log(2^2) = 2\log(2) = 2\Delta l\\ \log(8)&= \log(2^3) = 3\log(2) = 3\Delta l\\ \log(16) &= \log(2^4) = 4\log(2) = 4\Delta l \\ &\vdots \\ \log(2^N) &= N\log(2) = N\Delta l. \end{align}

So, with exponentially changing number of engines the noise level changes linearly! It is, in fact, true in general, for any \(a, b\) and \(N\): \[\log_b(a^N) = N\log_b(a).\]

Based on this simple observation we can also see that:

\begin{align}\log(4)&=\log(2\cdot 2) - \log(2) + \log(2) = 2\Delta l\\ \log(8)&=\log(2\cdot 4) = \log(2) + \log(4) = \Delta l + 2\Delta l = 3\Delta l\\ \log(16)&=\log(4 \cdot 4) = \log(4) + \log(4) = 2\Delta l + 2\Delta l = 4\Delta l.\end{align}

Which shows us how the multiplication of two number turns into sum of their logarithms. This is as well true in a more general case, for any \(a_1\) and \(a_2\):

\[\log_b(a_1\cdot a_2) = \log_b(a_1) + \log_b(a_2).\]

Note, that the above is by no means a proof, but rather a simple observation. Using those general relations we can conlude the following:

\begin{align}\log_b\left(\frac{a_1}{a_2}\right)&=\log_b(a_1\cdot a_2^{-1})\\ &= \log_b(a_1) + \log_b(a_2^{-1})\\ &=\log_b(a_1)-\log_b(a_2)\end{align}

for \(a_2 \neq 0.\) The latter fact is really the clue for introducing decibels. Before going further into that, let us explain what is the mysterious number \(b\) in \(\log_b(a)\). To this end, let us revise the definition \[\log_b(a)= c \iff b^c = a\]

The number \(b\) is called the base of the logarithm. As an example we could change the way we represent our number of engines as the exponents of 2. In such a case our base \(b\) would be equal to \(2\). We have:

\begin{align}\log_2(2) &=1 &\ \text{because}&\ \ 2^1 = 2\\ \log_2(4) &= 2 &\ \text{because}&\ \ 2^2 = 4\\ \log_2(8) &= 3&\ \text{because}&\ \ 2^3 = 8\\ \log_2(16)&= 4 &\ \text{because} &\ \ 2^4 = 16\\ \vdots \end{align}

We can therefore change the labeling in our plot. Can you see why?