Solving logs - Richter Scale and Decibels

  • #1
the text tells me that for calculating the Richter Scale magnitude of an earthquake we can use:

M = log(I/I0) which can also be written as

I = I0 x 10M

Where M=magnitute, I=intensity, and I0=intensity 0


How are those two formulas equal? Where did the log go? Can someone show me the proof for this?
 

Answers and Replies

  • #2
409
0
The log in the first equation must be to the base 10.

To cancel the log you do:

[itex]10^{M}=10^{Log(I/I_{0})}[/itex]

Which becomes:

[itex]10^{M}=I/I_{0}[/itex]

Then multiply by [itex]I_{0}[/itex]

[itex]I = 10^{M} * I_{0}[/itex]
 
  • #3
damn sorry the first one was supposed to be :

M = 10log(I/I0)

does that make a difference?
 
  • #4
33,722
5,418
damn sorry the first one was supposed to be :

M = 10log(I/I0)
That doesn't look right to me.

Starting with your 2nd formula,

I = I0 x 10M,
Divide both sides by I0 to get I/I0 = 10M

Now take the log (log10 or common log) of both sides.
 
  • #5
yes, I think I understand what to do once I have the formula I=I0x10M

my problem is how to get to that from: M = 10log(I/I0)

that is the formula they gave me.

30 = 10log(I/I0)
log(I/I0) = 3
I = I0 x 103

I have no idea how they figured out each step but that is what I was given and I want to know the proof for that.
 
  • #6
eumyang
Homework Helper
1,347
10
Do you know the relationship between exponents and logarithms?
log x = b iff 10b = x.

So, looking at your last post, I'll insert a step.
30 = 10log(I/I0)
log(I/I0) = 3
103 = I/I0
I = I0 x 103

Do you see it now?
 
  • #7
ooooooooh right wow.. how did I miss that? :P alright thanks!
 
  • #8
33,722
5,418
yes, I think I understand what to do once I have the formula I=I0x10M

my problem is how to get to that from: M = 10log(I/I0)
What I'm saying is that you can't get there from this formula. Here's why:
M = 10log(I/I0)
=> M/10 = log(I/I0)
=> 10M/10 = I/I0
=> I = I010M/10

This is different from the formula you show.

Are you sure you're not misreading what they gave you? Or whoever wrote that formula might have made a typo, and typed "10log" instead of "log10".
that is the formula they gave me.

30 = 10log(I/I0)
log(I/I0) = 3
I = I0 x 103

I have no idea how they figured out each step but that is what I was given and I want to know the proof for that.
 
  • #9
What I'm saying is that you can't get there from this formula. Here's why:
M = 10log(I/I0)
=> M/10 = log(I/I0)
=> 10M/10 = I/I0
=> I = I010M/10

This is different from the formula you show.

Are you sure you're not misreading what they gave you? Or whoever wrote that formula might have made a typo, and typed "10log" instead of "log10".
but thats exactly right? if you just sub in 30 for M it works out perfectly no?
 
  • #10
33,722
5,418
No, that's not exactly right. In post #9 I started with M = 10log(I/I0), solved for I, and got I = I010M/10.

Your formula from post #1 is I = I010M.

I hope that you can see that these are not the same.
 
  • #11
yeah I see what I did, it was just a mistake on the first post, but I get it now.

also, just from looking at this question I was wondering:

if M = log(I/I0) then I = I0 x 10M... would this be the same as... I = I0eM

'e' as in exp.
 
  • #12
237
5
if M = log(I/I0) then I = I0 x 10M... would this be the same as... I = I0eM

'e' as in exp.
No. 10 does not equal e, does it?

Logarithm to the base 10 is often denoted as "log"; logarithm to the base e is generally denoted as "ln".
 
  • #13
No. 10 does not equal e, does it?

Logarithm to the base 10 is often denoted as "log"; logarithm to the base e is generally denoted as "ln".
oh no sorry I meant 'E' as in scientific notation.
like 99E7 = 99x107 = 990,000,000

is that the same for the formula in my last post?
 
  • #14
237
5
oh no sorry I meant 'E' as in scientific notation.
like 99E7 = 99x107 = 990,000,000

is that the same for the formula in my last post?
Ah, yes. I've never seen it written that way before other than on a calculator, but I do know what you mean.
 
  • #15
33,722
5,418
oh no sorry I meant 'E' as in scientific notation.
like 99E7 = 99x107 = 990,000,000

is that the same for the formula in my last post?
They don't usually write the exponent as a superscript. With the E notation, it would be 99E7, or more likely, 9.9E8 or 9.9E08.
 
  • #16
ok right. But if M = log(I/I0) then I = I0 x 10M... would this be the same as... I = I0EM

do you see what I'm saying? Because it would be a lot easier just to take the Io value and multiply it by 10 to the M every time, in such a situation.
 
  • #17
eumyang
Homework Helper
1,347
10
ok right. But if M = log(I/I0) then I = I0 x 10M...
You mean I = I0 x 10M.

...would this be the same as... I = I0EM
I guess so, but it's not usually written that way... only in calculators/computers. What if you have an exponential expression and the base is not 10?
 
  • #18
33,722
5,418
ok right. But if M = log(I/I0) then I = I0 x 10M... would this be the same as... I = I0EM
You mean I = I0 x 10M.
do you see what I'm saying? Because it would be a lot easier just to take the Io value and multiply it by 10 to the M every time, in such a situation.
I guess so, but it's not usually written that way... only in calculators/computers. What if you have an exponential expression and the base is not 10?
I doubt that anyone would look at I0EM and comprehend that M is supposed to be the exponent on 10. Instead, most people would interpret this as I0 * E * M, where E and M would be presumed to be some unstated values. I have never seen scientific notation in programming form (i.e., E+nn form) where the exponent is a variable.
 

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