- #1

- 31

- 2

M = log(I/I

_{0}) which can also be written as

I = I

_{0}x 10

^{M}

Where M=magnitute, I=intensity, and I

_{0}=intensity 0

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

- Thread starter Gregory.gags
- Start date

- #1

- 31

- 2

M = log(I/I

I = I

Where M=magnitute, I=intensity, and I

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

- #2

- 409

- 0

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

- 31

- 2

damn sorry the first one was supposed to be :

M = 10log(I/I_{0})

does that make a difference?

M = 10log(I/I

does that make a difference?

- #4

Mark44

Mentor

- 33,722

- 5,418

That doesn't look right to me.damn sorry the first one was supposed to be :

M = 10log(I/I_{0})

Starting with your 2nd formula,

I = I

Divide both sides by I

Now take the log (log

- #5

- 31

- 2

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

that is the formula they gave me.

30 = 10log(I/I

log(I/I

I = I

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

log x = b iff 10

So, looking at your last post, I'll insert a step.

30 = 10log(I/I

log(I/I

10

I = I

Do you see it now?

- #7

- 31

- 2

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

- #8

Mark44

Mentor

- 33,722

- 5,418

What I'm saying is that you can't get there from this formula. Here's why:yes, I think I understand what to do once I have the formula I=I_{0}x10^{M}

my problem is how to get to that from: M = 10log(I/I_{0})

M = 10log(I/I

=> M/10 = log(I/I

=> 10

=> I = I

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 "log

that is the formula they gave me.

30 = 10log(I/I_{0})

log(I/I_{0}) = 3

I = I_{0}x 10^{3}

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

- 31

- 2

but thats exactly right? if you just sub in 30 for M it works out perfectly no?What I'm saying is that you can't get there from this formula. Here's why:

M = 10log(I/I_{0})

=> M/10 = log(I/I_{0})

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

=> I = I_{0}10^{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 "log_{10}".

- #10

Mark44

Mentor

- 33,722

- 5,418

Your formula from post #1 is I = I

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

- #11

- 31

- 2

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

if M = log(I/I

'e' as in exp.

- #12

- 237

- 5

No. 10 does not equal e, does it?if M = log(I/I_{0}) then I = I_{0}x 10^{M}... would this be the same as... I = I_{0}e^{M}

'e' as in exp.

Logarithm to the base 10 is often denoted as "log"; logarithm to the base e is generally denoted as "ln".

- #13

- 31

- 2

oh no sorry I meant '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".

like 99

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

- #14

- 237

- 5

Ah, yes. I've never seen it written that way before other than on a calculator, but I do know what you mean.oh no sorry I meant '_{E}' as in scientific notation.

like 99_{E}^{7}= 99x10^{7}= 990,000,000

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

- #15

Mark44

Mentor

- 33,722

- 5,418

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.oh no sorry I meant '_{E}' as in scientific notation.

like 99_{E}^{7}= 99x10^{7}= 990,000,000

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

- #16

- 31

- 2

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

You mean I = Iok right. But if M = log(I/I_{0}) then I = I_{0}x 10M...

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?...would this be the same as... I = I_{0}EM

- #18

Mark44

Mentor

- 33,722

- 5,418

ok right. But if M = log(I/I_{0}) then I = I_{0}x 10M... would this be the same as... I = I_{0}EM

You mean I = I_{0}x 10^{M}.

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 doubt that anyone would look at II 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?

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 4K

- Last Post

- Replies
- 2

- Views
- 964

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 11

- Views
- 3K