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I'm trying to understand the formula to move from log points to percentage points. I know the same question has already been asked here: How to interpret the difference in log points

and I can follow PaulB's answer easily until the taylor expansion, is the last step that I have troubles understanding. Could anyone please help me claryfying that? It seems to me like the "-1" should be part of the log, but it's clearly not correct.

Any help would be highly appreciated

Gabriele Dente
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  • The exponent part is just $$ e^{\ln x - \ln y} = e^{\ln x} \cdot e^{-\ln y} = x \cdot y^{-1} = \frac{x}{y} $$ Was that the confusing part? – Matti P. Aug 04 '20 at 10:20

1 Answers1

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We have that when x is "small" (let try with calculator)

$$\log(1+x)\approx x$$

therefore for small pecentage $\frac{\%\Delta}{100}$

$$\log\left(1+\frac{\%\Delta}{100}\right) \approx \frac{\%\Delta}{100}$$

For the last step, by exponentiation

$$\log(x)-\text{log}(y)=\log\left(\frac{\%\Delta}{100}+1\right)$$

$$e^{\log(x)-\text{log}(y)}=e^{\log\left(\frac{\%\Delta}{100}+1\right)}=\frac{\%\Delta}{100}+1$$

$$\frac{\%\Delta}{100}=e^{\log(x)-\text{log}(y)}-1$$

user
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  • Thank you, now I get it. It's just about rearranging. It was very stupid of me not to understand it before! Thank you so much :-) – Gabriele Dente Aug 04 '20 at 10:29
  • @GabrieleDente You are welcome! Glad to can help you with this little issue. Bye – user Aug 04 '20 at 10:31