# If the data appear to fit a power function, what might be the most appropriate non-linear transformation, to linearize the data?

Jun 26, 2015

#### Answer:

You could call it a "log-log transformation" (make a "log-log plot ")

#### Explanation:

If $y$ is a power function of $x$, then $y = a {x}^{n}$ for some constants $a$ and $n$. Taking the log of both sides (say, the common logarithm (base 10), but any log will do) gives

$\log \left(y\right) = \log \left(a {x}^{n}\right)$

Using properties of logarithms, this can be written as

$\log \left(y\right) = \log \left(a\right) + n \log \left(x\right)$

Letting $Y = \log \left(y\right)$, $X = \log \left(x\right)$, and $A = \log \left(a\right)$, this equation becomes

$Y = A + n X$, giving $Y = \log \left(y\right)$ as a linear function of $X = \log \left(x\right)$.

For example, suppose your data consisted of the points $\left(2 , 6.7\right)$, $\left(3 , 18.8\right)$, $\left(4 , 38.4\right)$, and $\left(5 , 66.9\right)$. Plotting these data gives a definite nonlinear trend in the graph shown below.

Suppose you suspect the relation between $x$ and $y$ is a power function. Take the log of both the $x$- and $y$-coordinates of your data, to get $X$- and $Y$-coordinates of data for a log-log plot: $\left(0.301 , 0.826\right) , \left(0.477 , 1.274\right) , \left(0.602 , 1.584\right) , \left(0.699 , 1.825\right)$. This plot has a definite linear trend.

In fact, if you find the least-squares linear regression line for this second graph, you'll get approximately $Y = 0.072499 + 2.5106 X$. This implies that $\log \left(y\right) = 0.072499 + 2.5106 \log \left(x\right)$ so that $y = {10}^{0.072499 + 2.5106 \log \left(x\right)} = {10}^{0.072499} \cdot {10}^{\log \left({x}^{2.5106}\right)} \setminus \approx 1.18168 {x}^{2.5106}$. The final graph shows that this is a good fit for the original $x y$-data.