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This equation:

$\hspace{1in}X=S\times(P_{FC}/P_{DC})$

Can be rewritten in terms of percentage change over a time period as follows:

$\hspace{1in}x=s+I_{FC}-I_{DC}=s-(I_{DC}-I_{FC})$

How can I actually rewrite this? I think it has something to do with logarithms because the multiplication becomes an addition and the division becomes a subtraction, but I think taking the log on both sides won't give me the answer.

For example,since $x$ is a percentage change over a time period, I figured its this must be true:

$\hspace{1in} x=\frac{X_{2}}{X_{1}}-1$

But $\frac{X_{2}}{X_{1}}-1$ is definitely not equal to $\log X$.

Any hints would be much appreciated. Thank you.

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Let $x = \log X, s = \log S, I_{FC} = \log P_{FC}, I_{DC} = \log P_{DC}$. –  PEV Dec 30 '10 at 15:13
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2 Answers

up vote 0 down vote accepted

The confusion stems from the fact that you do not say what quantities change. Say, if you mean that $X$ changes by a small amount $\Delta X$ then

$$x=\log (X+\Delta X) - \log X = \log (1+\frac{\Delta X}{X}) \approx \frac{\Delta X}{X} \; .$$

So $x$ is really the relative change in the quantity $X$ and likewise for the other quantities.

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So $X = S \times (P_{FC}/P_{DC})$.

Then $\log X = \log S + (\log P_{FC}- \log P_{DC})$.

Thus $\log X = \log S-(\log P_{DC}- \log P_{FC})$.

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