Transposing 20log(x) =-6 to find x

Thread Starter

dunc_0713

Joined Oct 27, 2017
3
Hi,

I have had a look in the forum to see if this question has come up before but I don't think it has.

I'm having trouble transposing this equation, 20*log(x) = -6, to find x or to just get x on it's own. I know that x = 0.5 in the equation in question but I'm having trouble proving how I got this mathematically.

Any help will be much appreciated!
 

neonstrobe

Joined May 15, 2009
73
Hi,

I have had a look in the forum to see if this question has come up before but I don't think it has.

I'm having trouble transposing this equation, 20*log(x) = -6, to find x or to just get x on it's own. I know that x = 0.5 in the equation in question but I'm having trouble proving how I got this mathematically.

Any help will be much appreciated!
You need anti-logs. That is 10^x for logs to base 10.
I'm sure you can divide both sides by 20.
 

Thread Starter

dunc_0713

Joined Oct 27, 2017
3
You need anti-logs. That is 10^x for logs to base 10.
I'm sure you can divide both sides by 20.
Thank you!

Works perfectly, and I understand how to do it now. Surprisingly we hadn't covered this in previous years of study so I had no idea.


Thanks again!
 

WBahn

Joined Mar 31, 2012
24,684
By definition, the base-b logarithm of x is the number y such that, when b is raised to the power y you get x.

\[
y \; = \; log_b(x)
x \; = \; b^y \; = \; b^{log_b(x)}
\]

NOTE: Something appears to be wrong with AAC's MimeTex rendering engine. So here it is in text format:

y = log_b(x)

x = b^y = b^log_b(x)

In your case here (given that we know where the relation is coming from) the base is 10.

Sadly, it is increasingly common that people are being "taught" logs and exponentials only in terms of which buttons to punch on a calculator to evaluate them without being taught what they are, what they mean, or their inverse relationship to each other.
 
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Thread Starter

dunc_0713

Joined Oct 27, 2017
3
By definition, the base-b logarithm of x is the number y such that, when b is raised to the power y you get x.

\[
y \; = \; log_b(x)
x \; = \; b^y \; = \; b^{log_b(x)}
\]

NOTE: Something appears to be wrong with AAC's MimeTex rendering engine. So here it is in text format:

y = log_b(x)

x = b^y = b^log_b(x)

In your case here (given that we know where the relation is coming from) the base is 10.

Sadly, it is increasingly common that people are being "taught" logs and exponentials only in terms of which buttons to punch on a calculator to evaluate them without being taught what they are, what they mean, or their inverse relationship to each other.
Yes you are absolutely right! I must admit that while I have been taught a little about exponentials and logarithms the past few years, it has mainly just been in terms of what to put into the calculator.
 

WBahn

Joined Mar 31, 2012
24,684
Yes you are absolutely right! I must admit that while I have been taught a little about exponentials and logarithms the past few years, it has mainly just been in terms of what to put into the calculator.
I'm not surprised. I wish I were. About a decade ago I was involved in an after-class discussion with the two students whom I had already pegged as the brightest two students in the class (and, as it turned out, they had both been their high school class valedictorians). They were both completely confused by how I went from the expression

\[
\ln(z) \; = \; y \cdot \ln(x)
\]

to

\[
z \; = \; e^{y \cdot \ln(x)}
\]

As soon as I pointed out that the definition of a logarithm is that

\[
z \; = \; b^{log_b(z)}
\]

they understood immediately.

So we got to talking and it turned out that both of them had been taught logarithms and exponentials, but in both cases they had been told that sometimes you will see expressions such as ln(something) or e^something and, when you do, this is the button you press on your TI-83 calculator to evaluate it. They hadn't been taught ANYTHING further. Normally I would have been suspicious that they really had, but it had just gone in one ear and out the other. But not with these two -- they were genuinely curious about understanding the details of what they were learning, but exponentials and logs had been presented in such a dissociated way that they had no idea that there was anything deeper to ask.

I've since learned that many trig classes now do the same thing and have even seen high school trig books that "explain" the sine function solely by showing a picture of a TI-83 (I think that's what it was) calculator with a circle around the sin(x) button. No where in the text could I find a definition of the sine of an angle in terms of the sides of a right triangle. Maybe it was there, but if it was it was in there as little more than a side note tucked away somewhere.
 

hobbyist

Joined Aug 10, 2008
887
Sadly, it is increasingly common that people are being "taught" logs and exponentials only in terms of which buttons to punch on a calculator to evaluate them without being taught what they are, what they mean, or their inverse relationship to each other.
Very well said!
 

neonstrobe

Joined May 15, 2009
73
... and the term "anti-log" has not been mentioned I think since log tables got canned by the calculator. Sometimes I wonder whether it would be useful to teach the old school stuff like log tables and slide rules... just to show how these things worked, not for serious use of course. I still put two rulers together to explain adding and subtracting to junior level students, then go on to discuss logs and exponentials.
 

MrAl

Joined Jun 17, 2014
6,459
Hi,

I have had a look in the forum to see if this question has come up before but I don't think it has.

I'm having trouble transposing this equation, 20*log(x) = -6, to find x or to just get x on it's own. I know that x = 0.5 in the equation in question but I'm having trouble proving how I got this mathematically.

Any help will be much appreciated!
Hi,

This probably doesnt matter as much for this problem but for:
20*log(x)=-6

The value of x to an approximation of 8 digits is:
0.50118723

and x=0.5 would lead to
20*log(0.5)=-6.0205999

approximately.

This is sometimes important, sometimes not important. For example in a filter response usually stating -6db is close enough.

As a side note, because of automatic math software these days we should mention whether or not log(x) is the base 10 or base 'e' log. In the past log(x) was alwaays meant to mean log base 10 of x and ln(x) was the natural log, but in recent years log(x) now sometimes (and quite often now) means the natural log which is really ln(x) so log(x)=ln(x) in those cases. I try to remember to state this when i reply to something.
In this case it was a little more apparent though from the context.
If log(x) was really the natural log here, we would get:
20*log(0.5)=-13.862944

approximately, or when solving:
20*log(x)=-6

we would get:
x=0.74081822

approximately.
 
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