forming log equation from graph points

Question

Okay so I need to form a logarithmic equation from the points (1960,4.7) (1964,5.1) (1968,5.4). I have 'guess and checked' to get the equation 2.7421 log(x-1950)+1.9579, and was wondering if there was a better way to get to that equation, or if there is a more exact equation/way to find a more exact equation that isn't too complicated? Its only for year 11 maths, so please no 3rd year engineering equations aha. Thank you in advance and I need this question finished in three days so responses would be great!

Answer 1

It depends very much on the model you select. With more points than parameters, the problem is related to linear or non linear regression.

If you want a model such $$y=a \log_{10}(x-b)+c$$ since you have three data points and three parameters, you can solve exactly three equations for the three unknown parameters $a,b,c$. So the equations to solve are $$4.7=a \log_{10}(1960-b)+c\tag 1$$ $$5.1=a \log_{10}(1964-b)+c\tag 2$$ $$5.4=a \log_{10}(1968-b)+c\tag 3$$ From $(1)$, you can extract $c$ as a function of $a,b$ $$c=4.7-a \log_{10}(1960-b)\tag 4$$ Plug it in $(2)$ and $(3)$ and group terms to have $$5.1=a \log_{10}(1964-b)+4.7-a \log_{10}(1960-b)$$ $$5.4=a \log_{10}(1968-b)+4.7-a \log_{10}(1960-b)$$ that is to say $$0.4=a\log_{10}\Big(\frac {1964-b}{1960-b}\Big)\tag 5$$ $$0.7=a\log_{10}\Big(\frac {1968-b}{1960-b}\Big)\tag 6$$ Make the ratio of the last equations to get $$\frac 47=\frac{\log_{10} \left(\frac{1964-b}{1960-b}\right)}{\log_{10} \left(\frac{1968-b}{1960-b}\right)}\tag 7$$ which is a quite nasty equation in $b$ but which can be solved almost exactly using numerical methods; for six significant figures, the solution is $b=1949.81$ so beatifully close to your $1950$.

Going backwards, you would get $a=2.78208$ and $c=1.89488$ which reproduce exactly the three data points.

You model was indeed very good but, recomputing the values, you would obtain $4.70000$, $5.10070$ and $5.39998$ which are indeed very good but not strictly exact.

You could notice that the parameters changed quite significantly.

Edit

We can rewrite the equation $(7)$ to be solved as $$\frac{(1964-b)^7}{(1960-b)^7}=\frac{(1968-b)^4}{(1960-b)^4}\implies (1964-b)^7=(1960-b)^3 (1968-b)^4$$ Let $b=B+1964$ and expand to get $$B^6+12 B^5-48 B^4-192 B^3+768 B^2+1024 B-4096=0\tag 8$$ and, because of the logarithms, $b < 1960 \implies B<-4$. The sextic polynomial cannot factor; it shows four comples roots, one real positive root and one real negative root the one of interest). Numerically $B=-14.1926\implies b=1949.8074$