Solving logs - Richter Scale and Decibels

Gregory.gags

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 10^M

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?

rollcast

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]

Gregory.gags

damn sorry the first one was supposed to be :

M = 10log(I/I₀)

does that make a difference?

Mark44

That doesn't look right to me.

Starting with your 2nd formula,

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

Now take the log (log₁₀ or common log) of both sides.

Gregory.gags

yes, I think I understand what to do once I have the formula I=I₀x10^M

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

Do you know the relationship between exponents and logarithms?
log x = b iff 10^b = 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?

Gregory.gags

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

Mark44

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)
=> 10^M/10 = I/I₀
=> I = I₀10^M/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".

Gregory.gags

but thats exactly right? if you just sub in 30 for M it works out perfectly no?

Mark44

No, that's not exactly right. In post #9 I started with M = 10log(I/I₀), solved for I, and got I = I₀10^M/10.

Your formula from post #1 is I = I₀10^M.

I hope that you can see that these are not the same.

Gregory.gags

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 10^M... would this be the same as... I = I₀e^M

'e' as in exp.

oay

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".

Gregory.gags

oh no sorry I meant '_E' as in scientific notation.
like 99_E⁷ = 99x10⁷ = 990,000,000

is that the same for the formula in my last post?

oay

Ah, yes. I've never seen it written that way before other than on a calculator, but I do know what you mean.

Mark44

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.

Gregory.gags

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

You mean I = I₀ x 10^M.

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?