I have several log(y) ~ a + b * log(x) models that I want to express as exponential relationships. I know this involves an expoential transformation, but how do I solve it?
Example:
Step 1: log(y) ~ 2 + 3 * log(x)
Step 2: exp(log(y)) ~ exp(2 + 3 * log(x))
Step 3: exp(log(y)) ~ exp(2) + exp(3 * log(x))
what comes next for the exp(3 * log(x)) part?
I'm using a natural logarithm.
Some logarithm properties to have in mind
(NOTE: when i write $\log $ I am referring to the natural logarithm a.k.a $\ln , \log_e$ also $e^a=\exp(a)$)
back to your example
You were almost there !
$\log(y)=2+3\log(x) \implies e^{\log(y)}=e^{2+3\log(x)} \implies y=e^2 \cdot e^{3\log(x)} \implies y=e^2 \cdot e^{\log(x^3)} \implies y=e^2 \cdot x^3$
(also note that $e^{a+b} = e^a \cdot e^b $)
EDIT
after a bit of playing around with equations of the form $\log(y)=a+b \cdot \log(c) $ we can see the following
$\log(y)=a+b \cdot \log(c) \implies \log(y)=\log e^a + \log(c^b) \implies \log(y)=\log(e^a \cdot c^b) $
$\implies e^{\log(y)}=e^{\log(e^a \cdot c^b)} \implies y= e^a \cdot c^b$
and that is the general simplified form of those types of expressions in terms of $y$. Although I would not recommend just taking this and substituting in you $a,b,c$ values especially if you are still not really comfortable with those types of problems