In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings, like those shown in Figure \(\PageIndex\). 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 whereas the Japanese earthquake registered a 9.0.
Figure \(\PageIndex\): Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce).
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
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 \(10^x=500\), where \(x\) represents the difference in magnitudes on the Richter Scale. How would we solve for \(x\)?
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve \(10^x=500\). We know that \(^2=100\) and \(^3=1000\), so it is clear that \(x\) must be some value between 2 and 3, since \(y=^x\) is increasing. We can examine a graph, as in Figure \(\PageIndex\), to better estimate the solution.
Figure \(\PageIndex\)
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure \(\PageIndex\) passes the horizontal line test. The exponential function \(y=b^x\) is one-to-one, so its inverse, \(x=b^y\) is also a function. As is the case with all inverse functions, we simply interchange \(x\) and \(y\) and solve for \(y\) to find the inverse function. To represent \(y\) as a function of \(x\), we use a logarithmic function of the form \(y=<\log>_b(x)\). The base \(b\) logarithm of a number is the exponent by which we must raise \(b\) to get that number.
We read a logarithmic expression as, “The logarithm with base \(b\) of \(x\) is equal to \(y\),” or, simplified, “log base \(b\) of \(x\) is \(y\).” We can also say, “\(b\) raised to the power of \(y\) is \(x\),” 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 \(2^5=32\), we can write \(<\log>_232=5\). We read this as “log base \(2\) of \(32\) is \(5\).”
We can express the relationship between logarithmic form and its corresponding exponential form as follows:
\[\begin \log_b(x)=y\Leftrightarrow b^y=x, b> 0, b\neq 1 \end\]
Note that the base \(b\) is always positive.
Because logarithm is a function, it is most correctly written as \(\log_b(x)\), using parentheses to denote function evaluation, just as we would with \(f(x)\). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as \(\log_bx\). Note that many calculators require parentheses around the \(x\).
We can illustrate the notation of logarithms as follows:
Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means \(y=<\log>^b(x)\) and \(y=b^x\) are inverse functions.
A logarithm base \(b\) of a positive number \(x\) satisfies the following definition.
Also, since the logarithmic and exponential functions switch the \(x\) and \(y\) values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,
No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.
Write the following logarithmic equations in exponential form.
Solution
First, identify the values of \(b\), \(y\),and \(x\). Then, write the equation in the form \(b^y=x\).
Write the following logarithmic equations in exponential form.
\(<\log>_(1,000,000)=6\) is equivalent to \(^6=1,000,000\)
\(<\log>_5(25)=2\) is equivalent to \(5^2=25\)
To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base \(b\),exponent \(x\),and output \(y\). Then we write \(x=<\log>_b(y)\).
Write the following exponential equations in logarithmic form.
Solution
First, identify the values of \(b\), \(y\),and \(x\). Then, write the equation in the form \(x=<\log>_b(y)\).
Write the following exponential equations in logarithmic form.
\(3^2=9\) is equivalent to \(<\log>_3(9)=2\)
\(5^3=125\) is equivalent to \(<\log>_5(125)=3\)
\(2^=\dfrac\) is equivalent to \(<\log>_2 \left (\dfrac \right )=−1\)
Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider \(<\log>_28\). We ask, “To what exponent must \(2\) be raised in order to get 8?” Because we already know \(2^3=8\), it follows that \(<\log>_28=3\).
Now consider solving \(<\log>_749\) and \(<\log>_327\) mentally.
Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate \(\log_> \frac\) mentally.
Solve \(y=<\log>_4(64)\) without using a calculator.
Solution
First we rewrite the logarithm in exponential form: \(4^y=64\). Next, we ask, “To what exponent must \(4\) be raised in order to get \(64\)?”
