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 [latex]{10}^{x}=500[/latex], where xrepresents the difference in magnitudes on the Richter Scale. How would we solve forx?
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[/latex]. We know that [latex]{10}^{2}=100[/latex] and [latex]{10}^{3}=1000[/latex], so it is clear that xmust be some value between 2 and 3, since [latex]y={10}^{x}[/latex] is increasing. We can examine a graphto better estimate the solution.
Figure 2
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph abovepasses the horizontal line test. The exponential function [latex]y={b}^{x}[/latex] is one-to-one, so its inverse, [latex]x={b}^{y}[/latex] is also a function. As is the case with all inverse functions, we simply interchange xand yand solve for yto find the inverse function. To represent yas a function of x, we use a logarithmic function of the form [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex]. The base blogarithm of a number is the exponent by which we must raise bto get that number.
We read a logarithmic expression as, "The logarithm with base bof xis equal to y," or, simplified, "log base bof xis y." We can also say, "braised to the power of yis 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 [latex]{2}^{5}=32[/latex], we can write [latex]{\mathrm{log}}_{2}32=5[/latex]. 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:
[latex]{\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1[/latex]
Note that the base bis always positive.
Because logarithm is a function, it is most correctly written as [latex]{\mathrm{log}}_{b}\left(x\right)[/latex], using parentheses to denote function evaluation, just as we would with [latex]f\left(x\right)[/latex]. 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 [latex]{\mathrm{log}}_{b}x[/latex]. 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 [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex] and [latex]y={b}^{x}[/latex] are inverse functions.
A General Note: Definition of the Logarithmic Function
A logarithm base bof a positive number xsatisfies the following definition.
For [latex]x>0,b>0,b\ne 1[/latex],
[latex]y={\mathrm{log}}_{b}\left(x\right)\text{ is equivalent to }{b}^{y}=x[/latex]
where,
- we read [latex]{\mathrm{log}}_{b}\left(x\right)[/latex] as, "the logarithm with base bof x" or the "log base bof x."
- the logarithm yis the exponent to which bmust be raised to get x.
Also, since the logarithmic and exponential functions switch the xand yvalues, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,
- the domain of the logarithm function with base [latex]b \text{ is} \left(0,\infty \right)[/latex].
- the range of the logarithm function with base [latex]b \text{ is} \left(-\infty ,\infty \right)[/latex].
Q & A
Can we take the logarithm of a negative number?
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.
How To: Given an equation in logarithmic form [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex], convert it to exponential form.
- Examine the equation [latex]y={\mathrm{log}}_{b}x[/latex] and identify b, y, and x.
- Rewrite [latex]{\mathrm{log}}_{b}x=y[/latex] as [latex]{b}^{y}=x[/latex].
Example 1: Converting from Logarithmic Form to Exponential Form
Write the following logarithmic equations in exponential form.
- [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex]
- [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
Solution
First, identify the values of b,y, andx. Then, write the equation in the form [latex]{b}^{y}=x[/latex].
- [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex]
Here, [latex]b=6,y=\frac{1}{2},\text{and } x=\sqrt{6}[/latex]. Therefore, the equation [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] is equivalent to [latex]{6}^{\frac{1}{2}}=\sqrt{6}[/latex].
- [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
Here, b= 3, y= 2, and x= 9. Therefore, the equation [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] is equivalent to [latex]{3}^{2}=9[/latex].
Try It 1
Write the following logarithmic equations in exponential form.
a. [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex]
b. [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]
Solution
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- Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[emailprotected]..
