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In the answer by fmark in this question it is stated that:

We prefer natural logs (that is, logarithms base $e$) because, as described above, coefficients on the natural-log scale are directly interpretable as approximate proportional differences: with a coefficient of 0.06, a difference of 1 in $x$ corresponds to an approximate 6% difference in $y$, and so forth.

Just wondering, if I apply log10 to the dependent and independent variable(s) would the above not apply anymore?

Looking at this one could argue that the above only applies for -5% ... 5% anyway and ln and log are very similar around x = 1 ...

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  • $\begingroup$ $log_{10}$ and $ln$ only differ by a (constant) factor so if you apply it to the dependent and the independent variables then it doesn't make a difference except for the constant, see purplemath.com/modules/logrules5.htm. If you denote the factor by $f$, i.e. $f\cdot ln(x)=log_{10}(x)$ then if $ln(y)=a+b\cdot ln(x)$ is equivalent to $f\cdot ln(x)= f\cdot a + f\cdot b \cdot ln(x)$ is equivalent to $log_{10}(y)=f\cdot a + b\cdot log_{10}(x)$ $\endgroup$ – user83346 Jul 23 '17 at 18:17
  • $\begingroup$ so the fitted parameter of a log10'ed independent variable has the same as that of an ln'ed independent variable variable if the dependent variable was log10'ed and ln'ed respectively? $\endgroup$ – cs0815 Jul 23 '17 at 18:20
  • $\begingroup$ except for the constant, see my comment $\endgroup$ – user83346 Jul 23 '17 at 18:20
  • $\begingroup$ Thanks. I was aware of this transformation but I am not sure I understand correctly. I fit log10(y) = alpha * log10(x) or ln(y) = beta * ln(x) can alpha be interpreted in the same manner as beta for small changes in x? $\endgroup$ – cs0815 Jul 23 '17 at 18:25
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    $\begingroup$ in that case $\alpha$ and $\beta$ should be identical, you can easily try that in e.g. R $\endgroup$ – user83346 Jul 23 '17 at 18:29
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As $ln(x)$ and $log_{10}(x)$ only differ by a factor see e.g. this link it holds that $f \cdot ln(x) = log_{10}(x)$, so then it holds that, if $ln(y)=a+b \cdot ln(x)$ then also $f \cdot ln(y)=f \cdot a+f \cdot b \cdot ln(x)$ or $log_{10}(y)= f \cdot a + b \log_{10}(x)$, so the coeffient $b$ is exactly the same.

Some R code to illustrate that:

x<-1:100

y<-5*x

lm(log(y)~log(x)-1)

lm(log10(y)~log10(x)-1)
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  • $\begingroup$ Although the coefficient remains the same, its interpretation is subtly different: it is no longer an approximate percent change in response per unit change in the (log) regressor. A difference of $1$ in $\log_{10} x$ now corresponds to a difference of $b$ in $\log_{10} y$, which (for small $|b|$) is approximately a $100\times b\log(10)\% \approx 230\, b\%$ change in $y$. $\endgroup$ – whuber Jul 24 '17 at 15:39
  • $\begingroup$ @whuber: I think a 1% change in $x$ corresponds to a b% change in $y$ in both cases , or am I wrong ? $\endgroup$ – user83346 Jul 24 '17 at 16:10
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    $\begingroup$ Yes, that is correct--which is why I characterized the difference in interpretation as "subtle". Notice that a "difference of 1" in $\log_{10}(x)$is not a $100\%$ change in $x$. Somebody who incorrectly supposes, as the OP explicitly does, that "ln and log are very similar around x=1" might very well fall into the trap of not realizing this. $\endgroup$ – whuber Jul 24 '17 at 16:29

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