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I have a very simple question. I am confused about the interpretation of log differences. Here a simple example: $$\log(2)-\log(1)=.3010$$ With my present understanding, I would interpret the result as follows: the number $2$ is $30,10\%$ greater than $1,$ which is obviously false. Can anyone lead me to the right interpretation?


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up vote 6 down vote accepted

Rather, it means that $2$ is (approximately) $10^{0.3010}$ times $1$.

More generally, for any positive $x,y,$ we have $$\log(x)+\log(y)=\log(xy),$$ and for any positive $c,$ we have (assuming we're dealing with base-$10$ logarithms rather than natural logarithms) $$c=\log(10^c).$$ Hence, the following are equivalent: $$\log(a)-\log(b)=c\\\log(a)=c+\log(b)\\\log(a)=\log(10^c)+\log(b)\\\log(a)=\log(10^cb).$$ By the uniqueness property of logarithms, this is then equivalent to $a=10^cb.$ If dealing with natural logarithms instead, $\log(a)-\log(b)=c$ is equivalent to $a=e^cb.$

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For example here: they say: "First difference of LOG = percentage change". So, what's wrong here? – DatamineR Aug 19 '13 at 21:43
They go on to explain that this is an approximation only, and not a true equality. It is very close when the percentage change is small, but for larger percentage changes, it may differ greatly. – Cameron Buie Aug 19 '13 at 21:49
Taking their example, they say the percentage change is $\frac {Y(t)-Y(t-1)}{Y(t-1)}$ without making explicit that you need to multiply by $100$. Taking the log of that gives $\log(Y(t)-Y(t-1))-\log(Y(t-1)$ where they have lost the $Y(t-1)$ in the first term. It works when the change is small, relying on the fact that $\log (1+x) \approx x$-and (as the page says) they are using natural logs, not base $10$ logs. – Ross Millikan Aug 19 '13 at 22:07

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