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I have a log-log plot of my data (see below)

enter image description here

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

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Yes it is antilog of "natural log" i.e. $\ln(x)$. Sometimes software (and texts) writes log to really mean $\log_e$ i.e. $\ln(x)$, whose "antilog" is $e^x$. – coffeemath Dec 23 '12 at 15:41
i see, so my conversion is correct? – Harpal Dec 23 '12 at 16:38
Yes, and it looks like the printout has that equation, labelled as "Unlogged" $y=1.924x^{.7089}$. – coffeemath Dec 23 '12 at 18:51
+1: This is a fine question and my answer to your previous thread was corrected (really sorry for the confusion...). The answer is correct if both logarithms are $\ln$ but $10^{5.26}$ must be chosen instead if $\log$ means $\log_{10}$. – Raymond Manzoni Dec 25 '12 at 11:55
Thanks for the help, does someone want to post an answer so I can accept it? – Harpal Dec 25 '12 at 13:50
up vote 2 down vote accepted

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 $\log_{10}$.

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