Answer: \(b=0.8\), \(A\approx1.21\), \(y\approx2.91\), \(x\approx3.12\).

The key point is to rewrite the equation using logarithm laws or a suitable substitution.

Taking logarithms of \(y=Ax^b\) gives

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

So the gradient of the straight line is \(b\), and the intercept is \(\lg A\).

Using the two points,

Now use the point \((0.61,0.57)\):

\(0.57=\lg A+0.8(0.61).\)

So

\(\lg A=0.57-0.488=0.082.\)

Hence

\(A=10^{0.082}\approx1.21.\)

(b) When \(x=3\),

\(\lg y=0.082+0.8\lg3.\)

This gives

\(y\approx2.91.\)

(c) When \(y=3\),

\(\lg3=0.082+0.8\lg x.\)

Therefore

\(\lg x=\frac{\lg3-0.082}{0.8}.\)

So

\(x\approx3.12.\)