Question

In 2014, Chile experienced an intense earthquake with a magnitude of 8.2 on the Richter scale. In 2010, Haiti also experienced an intense earthquake that measured 7.0 on the Richter scale. Compare the intensities of the two earthquakes. Use a logarithmic model to solve. Round to the nearest whole number.

Answer 1

Answer:

The intensity of the earthquake in Chile was about 16 times the intensity of the earthquake in Haiti.

Step-by-step explanation:

Given:

magnitude of earthquake in Chile =  8.2

magnitude of earthquake in Haiti = 7.0

To find:

Compare the intensities of the two earthquakes

Solution:

The magnitude R of earthquake is measured by R = log I

R is basically the magnitude on Richter scale

I is the intensity of shock wave

For Chile:

given magnitude R of earthquake in Chile =  8.2

R = log I

8.2 = log I

We know that:

$y = log a_{x}$  is equivalent to: $x = a^{y}$  

$R = log I$

8.2 = log I becomes:

$I = 10^{8.2}$

So the intensity of the earthquake in Chile:

$I_{Chile} = 10^{8.2}$

For Haiti:

R = log I

7.0 = log I

We know that:

$y = log a_{x}$  is equivalent to: $x = a^{y}$  

$R = log I$

7.0 = log I becomes:

$I = 10^{7.0}$

So the intensity of the earthquake in Haiti:

$I_{Haiti} = 10^{7}$

Compare the two intensities :

$\frac{I_{Chile} }{I_{Haiti} } }$

$= \frac{10^{8.2} }{10^{7} }$

= $10^{8.2-7.0}$

= $10^{1.2}$

= 15.848932

Round to the nearest whole number:

16

Hence former earthquake was 16 times as intense as the latter earthquake.

Another way to compare intensities:

Find the ratio of the intensities i.e. $\frac{I_{Chile} }{I_{Haiti} } }$

$log I_{Chile}$ - $log I_{Haiti}$ = 8.2 - 7.0

$log(\frac{I_{Chile} }{I_{Haiti} } })$ = 1.2

Convert this logarithmic equation to an exponential equation

$log(\frac{I_{Chile} }{I_{Haiti} } })$ = 1.2

$10^{1.2}$ = $\frac{I_{Chile} }{I_{Haiti} } }$

Hence

$\frac{I_{Chile} }{I_{Haiti} } }$  = 16