Question

How many times more intense is an earthquake that measures 7.5 on the Richter scale than one that measures 6.7? (Recall that the Richter scale defines magnitude of an earthquake with the equation M=log i/s , where i is the intensity of the earthquake being measured, and S is the intensity of a standard earthquake

Answer 1

6.31 times ( approx )

Step-by-step explanation:

Given formula to find the magnitude of an earthquake is,

`M=log((I)/(S))`

Where,

I = intensity of the earthquake being measured,

S = the intensity of a standard earthquake,

By the above formula,

`10^(M)=(I)/(S)`

`\implies I = S* 10^(M)`

Since, S is constant,

If M = 7.5,

Then intensity of the earthquake,

`I_1=S* 10^(7.5)`

If M = 6.7,

Then intensity of the earthquake,

`I_2=S* 10^(6.7)`

`(I_1)/(I_2)=(S* 10^(7.5))/(S* 10^(6.7))=10^(7.5-6.7)=10^(0.8)\approx 6.31`

`\implies I_1=6.31I_2`

Hence, the earthquake that measures 7.5 on the Richter scale is 6.3 times more intense than the earthquake that measures 6.7.

Answer 2

.5 - 6.7 = 0.8
10^0.8 ≅ 6.309573445
The 7.5 earthquake is approximately 6.3 times as intense as the 6.7 earthquake

How to figure it using your equation:
M = log(i/s)
10^M = i/s
i = s·10^M
ratio of the intensities of the two earthquakes = (s·10^7.5)/(s·10^6.7)
= 10^(7.5-6.7)
= 10^0.8
≅ 6.3:1