How to correctly transform a log-log graph into untransformed exponential graph?

Question

I plotted my data on a natural log-log scale and I seem to get a okay fit to the data with y=1.19 - 0.116x with Rsq = 0.29

I want to use the parameters but plot the row data with an exponential curve. Using my knowledge of exponents, I exponentiated both sides and tried to plot a curve of y ~ (-1.12 x) + 3.2 ...but I did not get a fit. I played around with some more functions, and the only fit I could get to work was this

    ggplot(data=df,aes(x=x.number,y=y.size)) + geom_point() + 
           stat_smooth(method="nls", formula =  y~(a*exp(-x*b) + 
           c),method.args=list(start=c(a=10, b=0.05, c=3)), se=F, 
           color="red") + stat_smooth(method="lm", formula =  
           y~a*exp(-x*b) + c , se=F, color="blue")

The formulae seems to require additional terms and the starting values are vastly different. I am trying to reconcile with the fits and I'm not sure how to go about it

Answer 1

For this response, I will the base-2 logarithm. If you obtained the regression parameters

• slope: -0.116
• intercept: 1.19

after transforming $(x,y)$ to $(\log_2(x),\log_2(y))$, then we have the following model $$\log_2(y) = 1.19 - 0.116\log_2(x) + \epsilon$$ which can be rewritten as $$y = \frac{2^{1.19}\cdot2^\epsilon}{x^{0.116}}$$

The log-log transform is often called the power model because it estimates a power-relationship between the $x$ and $y$ variables.

With regards to fitting the untransformed data to the original points, you must take into account the $2^\epsilon$ factor which models the error of the estimation. In the linear regression, the residual standard error gives the spread of the error term, the distribution of $\epsilon$. It is centered at zero (which transforms to a multiplicative factor of $2^0=1$), and we assume a normal distribution. This means the "error" scaling factor would range from $2^{-2\epsilon}$ and $2^{2\epsilon}$ (for roughly the "middlemost" 95% of the values).