Question

The Richter magnitude M is given by the model m=log(I/I_0) where I is the intensity of the earthquake in 100 km from the epicenter and I_0 is the smallest seismic activity that can be measured. A recent earthquake measured 6.2 on the Richter scale. How many times more intense was this earthquake than an earthquake that measured 5.8 on the Richter scale?

2.5 times more intense


5.5 times more intense


2.9 times more intense


3 times more intense

Answer 1

2.5 times more intense.

Explanation:

For the 6.2 magnitude earthquake, the Richter model gives us

`$6.2 = log((I)/(I_0) )$`

where
`I` is the intensity of the 6.2 earthquake.

And for the 5.8 magnitude earthquake,

`$5.8 = log((I_1)/(I_0) )$`

where
`I_1` is the intensity of the 5.8 earthquake.

Now, we subtract the two equations to get:

`$6.2 -5.8 = log((I)/(I_0) ) - log ((I_1)/(I_0) )$`

`$0.4 = log((I)/(I_0) ) - log ((I_1)/(I_0) )$`

Now using the logarithmic property

`$log (a)-log(b)= log((a)/(b) )$`

our equation becomes

`$0.4 = log((I)/(I_0) / (I_1)/(I_0) )$`

`$0.4 = log((I)/(I_0) * (I_0)/(I_1) )$`

`$0.4 = log((I)/(I_1) )$`

take both sides to the 10th power and get:

`$10^(0.4) =(I)/(I_1) $`

`$\boxed{(I)/(I_1) = 2.5} $`

Therefore, the earthquake was 2.5 times more intense.