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The magnitude M of an earthquake is measured using the formula M=log (x/c) , where x is the measured amplitude of a seismic wave and c is the reference amplitude of one micron. Two earthquakes are measured. The amplitude of a seismic wave in the first earthquake is 80 times the amplitude of a seismic wave in the second earthquake. How much greater is the magnitude of the first earthquake than the second earthquake?

Respuesta :

calculista

Answer:

The first earthquake is 1.9 times greater than the second

Step-by-step explanation:

We need to substract

M1 and M2

M1 - M2 = ㏒(x1/c) - ㏒(x2/c)

by properties of the logarithm

= log(x1) - log(c) - (log(x2) - log(c))

= log(x1) - log(x2)

But, we know that x1 = 80*x2

So,

= log(80) + log(x2) - log(x2)

= 1.903

MrRoyal

The magnitude of the first earthquake is 2 more than the second earthquake

The function is given as:

  • [tex]M = \log(\frac xc)}[/tex]

And the relationship between the amplitudes is given as:

  • [tex]x_1 = 80x_2[/tex]

Assume the magnitudes are M1 and M2.

So, the function can be rewritten as:

  • [tex]M_1 = \log(\frac{x_1}c)}[/tex]
  • [tex]M_2 = \log(\frac{x_2}c)}[/tex]

Subtract M2 from M1

[tex]M1 - M2 =\log(\frac{x_1}c)} - \log(\frac{x_2}c)}[/tex]

Apply the quotient property of the logarithm

[tex]M1 - M2 = log(x_1) - log(c) - (log(x_2) - log(c))[/tex]

Expand

[tex]M1 - M2 = log(x_1) - log(c) - log(x_2) + log(c)[/tex]

Evaluate the common terms

[tex]M1 - M2 = log(x_1) - log(x_2)[/tex]

Substitute [tex]x_1 = 80x_2[/tex]

[tex]M_1 - M_2 = log(80x_2) - log(x_2)[/tex]

Apply the product property of logarithm

[tex]M_1 - M_2 = log(80)+ log(x_2) - log(x_2)[/tex]

Evaluate the common terms

[tex]M_1 - M_2 = log(80)[/tex]

Evaluate log(80)

[tex]M_1 - M_2 = 1.9[/tex]

Approximate

[tex]M_1 - M_2 = 2[/tex]

Hence, the magnitude of the first earthquake is 2 more than the second earthquake

Read more about logarithmic functions at:

https://brainly.com/question/13473114

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