# what is 20 log (-10) =?

Discussion in 'Math' started by pot, Jan 1, 2016.

1. ### pot Thread Starter New Member

Nov 9, 2015
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what is 20 log (-10) =?

2. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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An error!

Log is not defined for negative numbers.

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Can you figure out a case where 10^x=-10?

4. ### pot Thread Starter New Member

Nov 9, 2015
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0

Must be briefed constant value even start drawing , use this equation ===> 20 Log(-10)

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Absolute value of magnitude?

Nov 23, 2012
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7. ### Wendy Moderator

Mar 24, 2008
20,772
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Going on a limb here, but it strikes me as an imaginary number, similar to the square root of -1.

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8. ### shteii01 AAC Fanatic!

Feb 19, 2010
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They are supposed to put frequency on the x-axis, and the absolute value of H(s) on the y-axis. Absolute value is Always positive. The correct formula is:
$
$

Last edited: Jan 2, 2016
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9. ### WBahn Moderator

Mar 31, 2012
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Your question lacks context. As a pure math question there are a couple of answers:

1) The normal logarithm function has the real numbers as both the domain and the range, hence the logarithm of a negative number is undefined since the base of normal logarithm functions are positive real numbers and there is no real exponent that you can apply to a positive real base and get a negative result.

2) The logarithm function can be extended to the complex numbers, in which the log of every non-zero argument is defined, though it is, in general, a multi-valued function.

But the look of the question implies that you are talking about voltage or current signals and converting them to decibels. In that case, you need to apply two functions, on for the magnitude and another for the phase. So you would have the magnitude as 20 log(10) and the phase as 180°.

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10. ### dannyf Well-Known Member

Sep 13, 2015
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Express -10 in the exponential form.

Alternatively, log(-10) = log(10*i^2) -> log (i). So express i in exponential form and you are done.

11. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Sure. In fact, there are infinitely many values of x for which 10^x yields -10. Here's one of them:

$
x \; = \; 1 + j \frac{\pi}{\ln(10)}
$

Let's check:

$
10^x \; = \; 10^{$$1 + j \frac{\pi}{\ln(10)}$$}
\;
10^x \; = \; $$10^1$$$$10^{\(j \frac{\pi}{\ln(10)}$$} \)
$

Since

$
10 \; = \; e^{\ln(10)}
$

we have

$
10^x \; = \; $$10^1$$$${\(e^{\ln(10)}$$}^{$$j \frac{\pi}{\ln(10)}$$} \)
\;
10^x \; = \; 10e^{$$j \frac{\pi \ln(10)}{\ln(10)}$$}
\;
10^x \; = \; 10e^{j \pi}
\;
10^x \; = \; 10$$-1$$
\;
10^x \; = \; -10
$

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