Why is the slope not back transformed in a regression equation for allometric relationships

I'm learning about allometric relationships and how to derive the parameters from regression equations.

I've seen that you can fit a linear regression model by taking the log of both the X and Y variables of your data that have an allometric relationship.

Then you can use the coefficients of the slope and intercept to create a power law of the form $Y=aX^b$

However, I've read that you need to raise your intercept to whatever base you're using but not your slope. For example, if I got an intercept of -1.2 after conducting a regression on the $log_{10}$ of both X and Y I need to apply $10^{-1.2}$ before using it in my power law.

My question is why don't you have to do this for the slope coefficient?

You have probably found estimates for $p$ and $q$ of the following model:

$$\log_{10}Y = p +q\log_{10}X$$

This is equivalent to

$$10^{\log_{10}Y} = 10^{p +q\log_{10}X} =10^{p +\log_{10}X^q}$$

$$\implies Y = 10^{p}X^q$$

Comparing with your model's original form,

$$Y=aX^b$$

you can see $a=10^p$ and $b=q$, i.e. you only take the power of your slope's estimate.

• Ah okay, so when I take the $log_{10}$ of my Y and X data I don't incorporate the coefficient estimate (i.e. the $\beta_1$ value which is the slope or q value in your answer). Rather it's because the slope is a coefficient of logged data in the first place that it's interpreted differently to the intercept. Jan 17 '17 at 22:36
• @Manassa Mauler, Yes, the slope you are finding is for the logged data, thats why you need the power for it to make sense for the original data. Jan 17 '17 at 22:46
• Thanks. I just found it confusing because you use typically use the slope to find the estimate of the intercept. Jan 17 '17 at 22:53