# what is 20 log (-10) =?

Thread Starter

#### pot

Joined Nov 9, 2015
9

#### djsfantasi

Joined Apr 11, 2010
7,693
An error!

Log is not defined for negative numbers.

#### shteii01

Joined Feb 19, 2010
4,644
Can you figure out a case where 10^x=-10?

#### shteii01

Joined Feb 19, 2010
4,644

#### GopherT

Joined Nov 23, 2012
8,012

#### Wendy

Joined Mar 24, 2008
22,436
Going on a limb here, but it strikes me as an imaginary number, similar to the square root of -1.

• atferrari

#### shteii01

Joined Feb 19, 2010
4,644
Going on a limb here, but it strikes me as an imaginary number, similar to the square root of -1.
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:
$$Magnitude\quad in \quad dB=20*log_{10}(|H(s)|)$$

Last edited:
• atferrari and pot

#### WBahn

Joined Mar 31, 2012
26,398
what is 20 log (-10) =?
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°.

• thumb2

#### dannyf

Joined Sep 13, 2015
2,197
what is 20 log (-10) =?
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.

#### WBahn

Joined Mar 31, 2012
26,398
Can you figure out a case where 10^x=-10?
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
\)

• thumb2 and Papabravo