Converting Between Logarithmic And Exponential Form
Learning Outcomes
- Convert from logarithmic to exponential form.
- Convert from exponential to logarithmic form.
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above passes 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 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 [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex]. The base b logarithm of a number is the exponent by which we must raise b to get that number.
tip for success
Understanding what a logarithm is requires understanding what an exponent is. A logarithm is an exponent. Read the paragraphs and boxes below carefully, perhaps more than once or twice, to gain the understanding of the inverse relationship between logarithms and exponents. Keep in mind that the inverse of a function effectively "undoes" what the other does. You can use the definition of the logarithm given below to solve certain equations involving exponents and logarithms.
Because a 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 when 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 b of a positive number x satisfies the following definition: For [latex]x>0,b>0,b\ne 1[/latex], [latex]y={\mathrm{log}}_{b}\left(x\right)\text{ is equal to }{b}^{y}=x[/latex], where- we read [latex]{\mathrm{log}}_{b}\left(x\right)[/latex] as, "the logarithm with base b of x" or the "log base b of x."
- the logarithm y is the exponent to which b must be raised to get x.
- if no base [latex]b[/latex] is indicated, the base of the logarithm is assumed to be [latex]10[/latex].
- 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: 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]
Answer: First, identify the values of b, y, and x. 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 equal 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 equal to [latex]{3}^{2}=9[/latex].
Try It
Write the following logarithmic equations in exponential form.- [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]
Answer:
- [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex] is equal to [latex]{10}^{6}=1,000,000[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex] is equal to [latex]{5}^{2}=25[/latex]
Convert from Exponential to Logarithmic Form
To convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base b, exponent x, and output y. Then we write [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].Example: Converting from Exponential Form to Logarithmic Form
Write the following exponential equations in logarithmic form.- [latex]{2}^{3}=8[/latex]
- [latex]{5}^{2}=25[/latex]
- [latex]{10}^{-4}=\frac{1}{10,000}[/latex]
Answer: First, identify the values of b, y, and x. Then, write the equation in the form [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].
- [latex]{2}^{3}=8[/latex] Here, b = 2, x = 3, and y = 8. Therefore, the equation [latex]{2}^{3}=8[/latex] is equal to [latex]{\mathrm{log}}_{2}\left(8\right)=3[/latex].
- [latex]{5}^{2}=25[/latex] Here, b = 5, x = 2, and y = 25. Therefore, the equation [latex]{5}^{2}=25[/latex] is equal to [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex].
- [latex]{10}^{-4}=\frac{1}{10,000}[/latex] Here, b = 10, x = –4, and [latex]y=\frac{1}{10,000}[/latex]. Therefore, the equation [latex]{10}^{-4}=\frac{1}{10,000}[/latex] is equal to [latex]{\text{log}}_{10}\left(\frac{1}{10,000}\right)=-4[/latex].
Try It
Write the following exponential equations in logarithmic form.- [latex]{3}^{2}=9[/latex]
- [latex]{5}^{3}=125[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex]
Answer:
- [latex]{3}^{2}=9[/latex] is equal to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
- [latex]{5}^{3}=125[/latex] is equal to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex] is equal to [latex]{\text{log}}_{2}\left(\frac{1}{2}\right)=-1[/latex]
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- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].