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

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

Answer: \(y=\ln\left(15.5-2x^3\right)\), valid for \(x\lt\sqrt[3]{\dfrac{31}{4}}\).

Let

\(X=x^3\quad\text{and}\quad Y=\mathrm{e}^y\).

The question says that plotting \(Y\) against \(X\) gives a straight line through \((1,13.5)\) and \((7.5,0.5)\).

The gradient of this straight line is

\(\dfrac{0.5-13.5}{7.5-1}=\dfrac{-13}{6.5}=-2\).

So the line has equation

\(Y=-2X+c\).

Using the point \((1,13.5)\),

\(13.5=-2(1)+c\),

so

\(c=15.5\).

Substituting back \(Y=\mathrm{e}^y\) and \(X=x^3\),

\(\mathrm{e}^y=15.5-2x^3\).

Taking natural logarithms gives

\(y=\ln\left(15.5-2x^3\right)\).

For this equation to be valid, the argument of the logarithm must be positive:

\(15.5-2x^3\gt0\).

Hence

\(x^3\lt7.75=\frac{31}{4}\).

Since the cube function is increasing, this is equivalent to

\(x\lt\sqrt[3]{\frac{31}{4}}\).

Answer: \(y=3\times5^x\), so \(A=3\) and \(b=5\).

If \(\ln y\) plotted against \(x\) gives a straight line, write

\(\ln y=mx+c.\)

The line passes through \((1,\ln15)\) and \((2,\ln75)\), so its gradient is

\(m=\dfrac{\ln75-\ln15}{2-1}.\)

Using the logarithm law \(\ln A-\ln B=\ln\left(\dfrac AB\right)\),

\(m=\ln\left(\dfrac{75}{15}\right)=\ln5.\)

Now use the point \((1,\ln15)\):

\(\ln15=\ln5+c.\)

Therefore

\(c=\ln15-\ln5=\ln3.\)

So

\(\ln y=x\ln5+ \ln3.\)

Exponentiating both sides gives

\(y=\mathrm e^{x\ln5+\ln3}=\mathrm e^{\ln3}\mathrm e^{x\ln5}.\)

Since \(\mathrm e^{\ln3}=3\) and \(\mathrm e^{x\ln5}=5^x\),

\(y=3\times5^x.\)

Hence

\(A=3\quad\text{and}\quad b=5.\)