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In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/#summary. Accessed 3/4/2013. . One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings, http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc0001xgp/#summary. Accessed 3/4/2013. like those shown in [link] . Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/. Accessed 3/4/2013. whereas the Japanese earthquake registered a 9.0. http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc0001xgp/#details. Accessed 3/4/2013.
The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is ${10}^{8-4}={10}^{4}=\mathrm{10,000}$ times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.
In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is $\text{\hspace{0.17em}}{10}^{x}=500,$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ represents the difference in magnitudes on the Richter Scale . How would we solve for $\text{\hspace{0.17em}}x?$
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve $\text{\hspace{0.17em}}{10}^{x}=500.\text{\hspace{0.17em}}$ We know that $\text{\hspace{0.17em}}{10}^{2}=100\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{10}^{3}=1000,$ so it is clear that $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ must be some value between 2 and 3, since $\text{\hspace{0.17em}}y={10}^{x}\text{\hspace{0.17em}}$ is increasing. We can examine a graph, as in [link] , to better estimate the solution.
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in [link] passes the horizontal line test. The exponential function $\text{\hspace{0.17em}}y={b}^{x}\text{\hspace{0.17em}}$ is one-to-one , so its inverse, $\text{\hspace{0.17em}}x={b}^{y}\text{\hspace{0.17em}}$ is also a function. As is the case with all inverse functions, we simply interchange $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ to find the inverse function. To represent $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x,$ we use a logarithmic function of the form $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right).\text{\hspace{0.17em}}$ The base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ logarithm of a number is the exponent by which we must raise $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ to get that number.
We read a logarithmic expression as, “The logarithm with base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is equal to $\text{\hspace{0.17em}}y,$ ” or, simplified, “log base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}y.$ ” We can also say, “ $b\text{\hspace{0.17em}}$ raised to the power of $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}x,$ ” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since $\text{\hspace{0.17em}}{2}^{5}=32,$ we can write $\text{\hspace{0.17em}}{\mathrm{log}}_{2}32=5.\text{\hspace{0.17em}}$ We read this as “log base 2 of 32 is 5.”
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