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I have the following example taken from the "ARRL General Class Manual", 7th Ed, 4-4.

Here's a practical application. Suppose you are using an antenna feed line that has a signal loss of 1dB. You can calculate the amount of transmitter power that's actually reaching our antenna and how much is lost in the feed line.

$$\text{Percentage Power} = 100\% \times \log^{-1}\left({-1\over10}\right) = 100\% \times \log^{-1}(-0.1) = 79.4\%$$

I think I'm missing something extremely trivial. In my calculator, I get

$$ 10^{-.1} = 9.9 $$

I don't have inverse log on my calculator but $ 10^{x} $ should be the same, no?

I think I'm missing the point with multiplying by 100% — that makes no sense to me. Either way I'm not sure how doing something to 9.9 with 100% gets me 79.4%.

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You have entered the formula in your calculator wrong. $10^{−0.1} ≈ 0.794$, as the book indicates. If you got $9.9$, then more likely you entered $10 - 0.1$ instead.

The 100% part is merely about explicitly converting multipliers to percentages, that is, $0.794 \times 100\% = 79.4$. (I don't care for that notation myself, and would rather define $\%$ such that $0.794 = 79.4\%$, period.) It's just a reminder that if you want a percentage value instead of a multiplier you have to multiply by 100.

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  • $\begingroup$ HAHA, Thanks, yes the classic hitting the subtract operation button instead of the +- button. I was checking work on previous examples but they were all positive values, so just my negative example wasn't working. Not sure why I didn't conclude 10-.01 = 9.9. Good thing I did not post this question to a public forum of global peers and that will most likely be around for a long time and is attached to my name :) $\endgroup$ – mikew Feb 5 '15 at 0:40

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