What is linear law?

Look at the straight line (in red). If I were to ask you to find the equation of the line, it would be a very easy task. You would only need to find the gradient and y-intercept, then plug them into the equation y=mx+c
However, if you had to find the equation of the curved line (in blue), it would be a difficult task. Even if I told you that its equation was $y=ax^n$, you still wouldn't be able to find out what a and n are easily.
Therefore, we convert (y=ax^n\) into a straight line. We then plot the straight line on a graph, determine the gradient and y-intercept, and use that to find x and n.

Converting non-linear equations into y=mx+c

(1) Exponentials
$ y = Ae^{kx} $
Example: $ y = (12p)e^{kx} $
$ \ln(y) = \ln[(12p)e^{kx}] $
$ \ln(y) = \ln(12p) + \ln(e^{kx}) $
$ \ln(y) = \ln(12p) + kx $
$ \ln(y) = kx + \ln(12p) $
$ Y = \ln(y) $
$ m = k $
$ X = x $
$ c = \ln(12p) $

(2) Power equations
$ y = ax^b $
Example: $ y = ax^{\frac{1}{2}} $
$ \log_{10}(y) = \log_{10}(ax^{\frac{1}{2}}) $
$ \log_{10}(y) = \log_{10}(a) + \log_{10}(x^{\frac{1}{2}}) $
$ \log_{10}(y) = \log_{10}(a) + \frac{1}{2}\log_{10}(x) $
$ \log_{10}(y) = \frac{1}{2}\log_{10}(x) + \log_{10}(a) $
$ Y = \log_{10}(y) $
$ m = \frac{1}{2} $
$ X = \log_{10}(x) $
$ c = \log_{10}(a) $

(3) Reciprocals
$ \frac{1}{y} = a + bx $
Example: $ \frac{1}{y} = x + a $
$ Y = \frac{1}{y} $
$ m = b $
$ X = x $
$ c = a $

(4) Power equation where x is the power
$ y = (a)(2k)^{bx} $
Example: $ y = (2k)^{3x} $
$ \log_{10}(y) = \log_{10}[(2k)^{3x}] $
$ \log_{10}(y) = 3x \log_{10}(2k) $
$ \log_{10}(y) = (\log_{10}(2k))3x $
$ Y = \log_{10}(y) $
$ m = \log_{10}(2k) $
$ X = 3x $
$ c = \log_{10}(2k) $

Linear law question example

Find $ n $ and $ k $.
$$PV^n = k$$

$$\begin{array}{|c|c|} \hline P & V \\ \hline 100 & 1.0 \\ \hline 80 & 1.2 \\ \hline 60 & 1.5 \\ \hline 40 & 2.0 \\ \hline \end{array}$$

$$\log_{10} PV^n = \log_{10} k$$
$$\log_{10} P + n \log_{10} V = \log_{10} k$$
$$\log_{10} P = -n \log_{10} V + \log_{10} k$$

$$\begin{array}{|c|c|} \hline \log_{10} P & \log_{10} V \\ \hline 2.00 & 0.00 \\ \hline 1.903 & 0.0792 \\ \hline 1.778 & 0.176 \\ \hline 1.602 & 0.301 \\ \hline \end{array}$$

We then plot a graph of $ \log_{10} P $ against $ \log_{10} V $.
From the graph, we find that:
$ -n $, gradient = -1.32
$ \log_{10} k $, y-intercept = 2.00
$ n = - \text{gradient} = -(-1.32) = 1.32 $
$ k = 10^{\text{y-intercept}} = 10^2 = 100 $