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Sagot :
To determine the correct equation for finding the amount of energy released by an earthquake with a magnitude of 2.7, let's solve the equation step-by-step.
The given equation is:
[tex]\[ M = \frac{2}{3} \log \frac{E}{E_0} \][/tex]
We are given:
[tex]\[ E_0 = 10^{4.4} \][/tex]
[tex]\[ M = 2.7 \][/tex]
We need to find [tex]\( E \)[/tex]. Follow these steps:
1. Substitute [tex]\( M \)[/tex] and [tex]\( E_0 \)[/tex] into the equation:
[tex]\[ 2.7 = \frac{2}{3} \log \frac{E}{10^{4.4}} \][/tex]
2. Multiply both sides by [tex]\(\frac{3}{2}\)[/tex] to isolate the logarithm:
[tex]\[ 2.7 \times \frac{3}{2} = \log \frac{E}{10^{4.4}} \][/tex]
[tex]\[ 4.05 = \log \frac{E}{10^{4.4}} \][/tex]
3. Recall that the logarithm operation [tex]\(\log\)[/tex] here is with base 10. To remove the logarithm, raise both sides as powers of 10:
[tex]\[ 10^{4.05} = \frac{E}{10^{4.4}} \][/tex]
Hence, the correct equation to use in order to find the amount of energy released ([tex]\(E\)[/tex]) is:
[tex]\[ 10^{4.05} = \frac{E}{10^{4.4}} \][/tex]
Therefore, the correct answer is:
[tex]\[ 10^{4.05} = \frac{E}{10^{4.4}} \][/tex]
The given equation is:
[tex]\[ M = \frac{2}{3} \log \frac{E}{E_0} \][/tex]
We are given:
[tex]\[ E_0 = 10^{4.4} \][/tex]
[tex]\[ M = 2.7 \][/tex]
We need to find [tex]\( E \)[/tex]. Follow these steps:
1. Substitute [tex]\( M \)[/tex] and [tex]\( E_0 \)[/tex] into the equation:
[tex]\[ 2.7 = \frac{2}{3} \log \frac{E}{10^{4.4}} \][/tex]
2. Multiply both sides by [tex]\(\frac{3}{2}\)[/tex] to isolate the logarithm:
[tex]\[ 2.7 \times \frac{3}{2} = \log \frac{E}{10^{4.4}} \][/tex]
[tex]\[ 4.05 = \log \frac{E}{10^{4.4}} \][/tex]
3. Recall that the logarithm operation [tex]\(\log\)[/tex] here is with base 10. To remove the logarithm, raise both sides as powers of 10:
[tex]\[ 10^{4.05} = \frac{E}{10^{4.4}} \][/tex]
Hence, the correct equation to use in order to find the amount of energy released ([tex]\(E\)[/tex]) is:
[tex]\[ 10^{4.05} = \frac{E}{10^{4.4}} \][/tex]
Therefore, the correct answer is:
[tex]\[ 10^{4.05} = \frac{E}{10^{4.4}} \][/tex]
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