Convert Between Logarithmic And Exponential Form
Learning Objectives
- Convert a logarithm to exponential form
- Convert an exponential expression to logarithmic form
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-display]y={\mathrm{log}}_{b}\left(x\right)\text{ is equivalent to }{b}^{y}=x[/latex-display] 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.
- 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 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
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 equivalent to [latex]{10}^{6}=1,000,000[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex] is equivalent to [latex]{5}^{2}=25[/latex]
Convert from exponential to logarithmic form
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 [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 equivalent 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 equivalent 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 equivalent 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 equivalent to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
- [latex]{5}^{3}=125[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex] is equivalent 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].