# Log + Ritcher Scale

• December 11th 2007, 04:35 PM
Macleef
Log + Ritcher Scale
Word Problem:
On September 26 2001, an earthquake in North Bay measured 5.0 on the Ritcher scale. What is the magnitude of an earthquake 3 times as intense as North Bay's earthquake?

My Work So Far:
Let $I_B$ represent the earthquake in North Bay
Let $I_F$ represent the earthquake in ________?

North Bay:
$5 = log(\frac {I_B}{I_o})$

$(\frac {I_B}{I_o}) = 10^{5}$

$I_B = 10^{5} I_o$

Somewhere Else:
$3r = log(\frac {I_F}{I_o})$

$(\frac {I_F}{I_o}) = 10^{3r}$

$I_F = 10^{3r} I_o$

Comparison:
$\frac {I_B}{I_F} = \frac {10^{5} I_o}{10^{3r} I_o}$

$\frac {I_B}{I_F} = \frac {10^{5}}{10^{3r}}$

${I_B}= 10^{5-3r}I_o$

Am I doing this right? If I am, what should I do next? If I'm not, where did I make my mistake and how should I correct it?

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• December 11th 2007, 09:52 PM
earboth
Quote:

Originally Posted by Macleef
Word Problem:
On September 26 2001, an earthquake in North Bay measured 5.0 on the Ritcher scale. What is the magnitude of an earthquake 3 times as intense as North Bay's earthquake?

My Work So Far:
Let $I_B$ represent the earthquake in North Bay
Let $I_F$ represent the earthquake in ________?

...

Hello,

you know that $I_F = 3 \cdot I_B$ . Thus you have:

$r_{I_F} = \log\left( \frac{I_F}{I_0}\right) = \log\left( \frac{3 \cdot I_B}{I_0} \right) =$ $\log\left(3 \cdot \frac{ I_B}{I_0} \right) = \log(3) + \log\left( \frac{ I_B}{I_0} \right) = \log(3) + 5 \approx 5.477121254...$