Choosing a Mathematical Model

First, plot the data on a graph as in Figure 8. For the purpose of graphing, round the data to two decimal places.

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try $y=a\ln (bx).$ Plugging in the first point, $\left(\text{1,0}\right)\text{,}$ gives $0=a\ln b.$ We reject the case that $a=0$ (if it were, all outputs would be 0), so we know $\ln (b)=0.$ Thus $b=1$ and $y=a\ln \left(\text{x}\right).$ Next we can use the point $\left(\text{9,4}\text{.394}\right)$ to solve for $a:$

Because $a=\frac{4.394}{\ln \left(9\right)}\approx 2,$ an appropriate model for the data is $y=2\ln \left(x\right).$

To check the accuracy of the model, we graph the function together with the given points as in Figure 9.

We can conclude that the model is a good fit to the data.

Compare Figure 9 to the graph of $y=\ln \left(x^2\right)$ shown in Figure 10.

The graphs appear to be identical when $x>0.$ A quick check confirms this conclusion: $y=\ln \left(x^2\right)=2\ln \left(x\right)$ for $x>0.$

However, if $x<0,$ the graph of $y=\ln \left(x^2\right)$ includes a “extra” branch, as shown in Figure 11. This occurs because, while $y=2\ln \left(x\right)$ cannot have negative values in the domain (as such values would force the argument to be negative), the function $y=\ln \left(x^2\right)$ can have negative domain values.