Question

The magnitude, M, of an earthquake is defined to be M=log I/S, where I is the intensity of the earthquake (measured by the amplitude of the seismograph wave) and S is the intensity of a “standard” earthquake, which is barely detectable. What is the magnitude of an earthquake that is 35 times more intense than a standard earthquake? Use a calculator. Round your answer to the nearest tenth.

Answer 1

log(35) = 1.544

 rounded to nearest tenth = 1.5

Answer 2

Answer: The magnitude of the earthquake that is 35 times more intense than a standard earthquake=1.5

Step-by-step explanation:

Given : The magnitude M of an earthquake is defined to be  $M=\log \frac{I}{S}$---(1)

, where I is the intensity of the earthquake (measured by the amplitude of the seismograph wave) and

S is the intensity of a “standard” earthquake, which is barely detectable.

If  the magnitude of an earthquake that is 35 times more intense than a standard earthquake

⇒Intensity of earthquake I=35 S

Substitute the value of I in (1), we get $M=\log(\frac{35S}{S})=\log(35)=\log(5\times7)=1.544068$ $\approx1.5$

∴ The magnitude of the earthquake that is 35 times more intense than a standard earthquake=1.5