1.17: Linear law: ln⁡y\ln y against x2x^{2}

0606/13/O/N/17 — Question 6 · 6 marks

When ln⁡y\ln y is plotted against x2x^{2} a straight line is obtained which passes through the points (0.2,2.4)(0.2, 2.4) and (0.8,0.9)(0.8, 0.9).
(i)
Express ln⁡y\ln y in the form px2+qpx^{2} + q, where pp and qq are constants.
[3]
(ii)
Hence express yy in terms of zz, where z=ex2z = e^{x^{2}}.
[3]