I have a log-log plot of my data (see below)
The equation of the line was determined to be: $5.26 + x0.7089$
If I wanted to convert this into a power equation would the correct way be:
$ ln(y) = A + B ln(x) $
Taking the antilog of both sides will give
y = $(e^A)(x^B)$
Let $e^A$ = A, and you have
$y = a(x^B)$
Is this the correct way to convert a line equation of log-log plot into a power equation in the form of $Ax^B$
If this is correct, why isnt the anti-log taken i.e. 10^5.26 instead of e^5.26
Best Answer
Supposing that both axis use b-basis logarithms we have : $$\log_b(y)=A+B\,\log_b(x)$$ using $\ b^{\log_b(y)}=y,\ b^{c+d}=b^c\,b^d\ $ and $\ B\,\log_b(x)=\log_b(x^B)\ $ we get : $$y=b^A\, x^B$$ So that your $\,y=e^A\,x^B$ equality was right for the natural logarithm $\,\ln=\log_e\,$ while the antilog expression ($y=10^{5.26}\,x^{0.7089}$) was the correct choice for the $10$-basis logarithm $\log_{10}$.