Answer: \(y=e^{-7}x^2\).

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

Since \(\ln y\) is plotted against \(\ln x\), write the straight line as

\(\ln y=m\ln x+c.\)

The line passes through \((6,5)\) and \((8,9)\), so its gradient is

\(m=\frac{9-5}{8-6}=2.\)

Using the point \((6,5)\),

\(5=2(6)+c.\)

Hence

\(c=-7.\)

Thus

\(\ln y=2\ln x-7.\)

Using \(2\ln x=\ln x^2\),

\(\ln y=\ln x^2-7.\)

Therefore

\(y=e^{-7}x^2.\)