If the question asked to sub y = 80, n = 1.25, a = 2.82 instead of estimation from the graph,

lg(80 - 12) = 1.25lgx + lg2.82

lg68 - lg2.82 = 1.25lgx

lg(68/2.82) = 1.25lgx

lgx = lg(68/2.82)/1.25

x = 10^(lg(68/2.82)/1.25)

x ≈ 12.7587416

x = 12.8 (3s.f)

If you substituted the values of a and n that were estimated from the graph directly into the equation, then lgx ≈ 1.103 and x would be 12.8 (3s.f)


But since the question asks you to estimate from the graph, then it's based on the accuracy of your drawing so your estimate of lg x = 1.0625 should be okay.

The graph fits the points better when lga ≈ 0.498 (based on computation using linear regression analysis, something that's beyond A Math syllabus).

But you're estimating from your drawing and due to scale limitations then your 0.45 from part (ii)(b) should be fine. Based on the scale, that's only about one small square away on the grid.

The gradient you obtained (n = 1.25) is very close to the regression fit of 1.25062.

In part (ii)(c), the value of lgx that you got is also close to the regression fit of 1.06635341887, which is good. We would get x = 11.65 (4s.f) which is close to your x = 11.5

In part (ii)(a) Regression fit gives the correct value of lg(y - 12) as 1.9697 which is close to your 1.975.

Overall, your graph is pretty accurate.