The magnitude, [tex]M[/tex], of an earthquake is defined to be [tex]M = \log \frac{I}{S}[/tex], where [tex]I[/tex] is the intensity of the earthquake (measured by the amplitude of the seismograph wave) and [tex]S[/tex] is the intensity of a "standard" earthquake, which is barely detectable.

What is the magnitude of an earthquake that is 1,000 times more intense than a standard earthquake? Use a calculator. Round your answer to the nearest tenth.

A. 2
B. 3
C. 4.5
D. 6.9



Answer :

To determine the magnitude [tex]\( M \)[/tex] of an earthquake that is 1,000 times more intense than a standard earthquake, follow these steps:

1. Identify the given values:
- The intensity of the earthquake, [tex]\( I \)[/tex], is 1,000 times more intense than a standard earthquake.
- The intensity of a standard earthquake, [tex]\( S \)[/tex], is the reference value.

Therefore, [tex]\( I = 1000 \times S \)[/tex].

2. Use the formula for the magnitude [tex]\( M \)[/tex]:
The magnitude [tex]\( M \)[/tex] is given by:
[tex]\[ M = \log_{10} \left( \frac{I}{S} \right) \][/tex]

3. Substitute the given values into the formula:
Since [tex]\( I = 1000 \times S \)[/tex], we can substitute this into the formula:
[tex]\[ M = \log_{10} \left( \frac{1000 \times S}{S} \right) \][/tex]

4. Simplify the fraction:
[tex]\[ M = \log_{10} \left( 1000 \right) \][/tex]

5. Calculate the logarithm:
[tex]\[ M = \log_{10} 1000 \][/tex]
We know that [tex]\( 10^3 = 1000 \)[/tex], so:
[tex]\[ \log_{10} 1000 = 3 \][/tex]

6. Round the answer to the nearest tenth:
Since 3 is already an integer, it can be represented as 3.0 when rounded to the nearest tenth.

Therefore, the magnitude of the earthquake is:
[tex]\[ M = 3.0 \][/tex]

So, the correct answer is:
[tex]\[ 3 \][/tex]

Other Questions