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The magnitude, [tex]M[/tex], of an earthquake is defined to be [tex]M = \log \left(\frac{I}{S}\right)[/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 35 times more intense than a standard earthquake? Use a calculator. Round your answer to the nearest tenth.

A. [tex]-1.5[/tex]
B. [tex]-0.5[/tex]
C. [tex]1.5[/tex]
D. [tex]3.6[/tex]

Sagot :

GinnyAnsweridn
To determine the magnitude, [tex]\( M \)[/tex], of an earthquake that is 35 times more intense than a standard earthquake, we use the provided formula:
[tex]\[ M = \log \frac{I}{S} \][/tex]

Here:
- [tex]\( I \)[/tex] is the intensity of the earthquake, which is 35 times that of the standard earthquake.
- [tex]\( S \)[/tex] is the intensity of a "standard" earthquake.

Substituting the given values into the formula, we have:
[tex]\[ M = \log \frac{35}{1} \][/tex]
[tex]\[ M = \log 35 \][/tex]

Using a calculator to find the logarithm base 10 of 35:
[tex]\[ \log 35 \approx 1.544068 \][/tex]

Rounding 1.544068 to the nearest tenth, we get:
[tex]\[ M \approx 1.5 \][/tex]

Therefore, the magnitude of the earthquake is:
[tex]\[ M \approx 1.5 \][/tex]

So, the correct answer is:
[tex]\[ 1.5 \][/tex]
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