# Solving logs - Richter Scale and Decibels

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?

Related Precalculus Mathematics Homework Help News on Phys.org
The log in the first equation must be to the base 10.

To cancel the log you do:

$10^{M}=10^{Log(I/I_{0})}$

Which becomes:

$10^{M}=I/I_{0}$

Then multiply by $I_{0}$

$I = 10^{M} * I_{0}$

damn sorry the first one was supposed to be :

M = 10log(I/I0)

does that make a difference?

Mark44
Mentor
damn sorry the first one was supposed to be :

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

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.

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.

eumyang
Homework Helper
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?

ooooooooh right wow.. how did I miss that? :P alright thanks!

Mark44
Mentor
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.

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?

Mark44
Mentor
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.

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.

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".

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?

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.

Mark44
Mentor
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.

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.

eumyang
Homework Helper
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?

Mark44
Mentor
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.