Answer: \(b\approx0.50\) and \(A\approx2.12\).

The relationship is

\(y=Ax^b\).

Take natural logarithms of both sides:

\(\ln y=\ln(Ax^b)\).

Using logarithm laws,

\(\ln y=\ln A+b\ln x\).

This has the straight-line form

\(Y=c+bX\),

where

\(Y=\ln y\), \(X=\ln x\), and \(c=\ln A\).

So on the graph of \(\ln y\) against \(\ln x\), the gradient is \(b\), and the vertical intercept is \(\ln A\).

The table values give approximately:

Using two points on the straight line, for example \((-0.50,0.50)\) and \((3.50,2.50)\),

\(b=\dfrac{2.50-0.50}{3.50-(-0.50)}=\dfrac{2.00}{4.00}=0.50\).

Now use

\(\ln y=\ln A+b\ln x\).

Substitute \((\ln x,\ln y)=(-0.50,0.50)\) and \(b=0.50\):

\(0.50=\ln A+0.50(-0.50)\).

Thus

\(0.50=\ln A-0.25\),

so

\(\ln A=0.75\).

Therefore

\(A=\mathrm{e}^{0.75}=2.117\ldots\).

Hence

\(A\approx2.12\quad\text{and}\quad b\approx0.50\).