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The magnitude $M$ of an earthquake is measured using the formula $M=\log(\frac{x}{c})$ , 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 15 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? Round your answer to the nearest hundredth.

Answer :

fichoh

Answer:

1.18

Step-by-step explanation:

Given the Magnitude of an earth quake model :

M = log(x/c)

x = measured amplitude ; c = reference amplitude

Amplitude of seismic wave in first earthquake = 15 amplitude of seismic wave in second

Let amplitude of seismic wave in first earthquake = x1

amplitude of seismic wave in second earthquake = x2

x1 = 15 * x2 = 15x2

Hence ;

1st earthquake magnitude :

M1 = log(a/c)

2nd earthquake magnitude :

M2 = log(15a/c)

M1 - M2

Log(x1/c) - log(x2/c)

Using the quotient rule

Log(x1) - log(c) - [log(x2) - log(c)]

Log(x1) - log(c) - log(x2) + log(c)

Log(x1) - log(x2)

Substituting x1 = 15x2

Log(x1) = log(15x2)

Seperating using product rule:

Log(x1) = log(15*x2) = log(15) + log(x2)

Hence,

log(15) + log(x2) - log(x2)

Log(15) - 0

= 1.1760

= 1.18 (nearest hundredth)

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