6.15: Solving Logarithmic Equations
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definitions and Theorems
A logarithmic equation is solved by rewriting it so that the variable is no longer trapped inside a logarithm. Two tools make this possible: converting a logarithm to exponential form, and the laws of logarithms.
A logarithmic equation is an equation in which a variable expression appears inside one or more logarithms.
Let \(b\) be a positive real number with \(b \neq 1\), and let \(x\) be a positive real number. The logarithmic equation\[\log_{b}(x)=y\nonumber\]is equivalent to the exponential equation\[b^{y}=x.\nonumber\]Rewriting a logarithm as the second equation is called converting to exponential form.
Let \(b\) be a positive real number with \(b \neq 1\), let \(m\) and \(n\) be positive real numbers, and let \(p\) be any real number. Then\[\log_{b}(mn)=\log_{b}(m)+\log_{b}(n),\nonumber\]\[\log_{b}\!\left(\dfrac{m}{n}\right)=\log_{b}(m)-\log_{b}(n),\nonumber\]\[\log_{b}(m^{p})=p\log_{b}(m).\nonumber\]These are called the product law, the quotient law, and the power law, respectively.
Let \(b\) be a positive real number with \(b \neq 1\), and let \(m\) and \(n\) be positive real numbers. Then\[\log_{b}(m)=\log_{b}(n) \quad \text{if and only if} \quad m=n.\nonumber\]
Before converting to exponential form, isolate a single logarithm on one side of the equation. If several logarithmic terms appear on the same side, first use the Laws of Logarithms to condense them into one logarithm.
The argument of a logarithm must be positive. Condensing logarithms and converting to exponential form can introduce values that violate this requirement. Substitute every proposed solution back into the original equation, and discard any value that makes the argument of a logarithm zero or negative. Such a value is extraneous.
Examples
Solve \(\log_{2}(3x-5)=4\).
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The equation contains a single logarithm, so convert directly to exponential form. Reading \(\log_{2}(3x-5)=4\) in exponential form gives\[2^{4}=3x-5.\nonumber\]Since \(2^{4}=16\),\[16=3x-5 \quad\Longrightarrow\quad 21=3x \quad\Longrightarrow\quad x=7.\nonumber\]The argument at \(x=7\) is \(3(7)-5=16>0\), so the value lies in the domain. The solution is \(x=7\).
Solve \(\ln(x-1)=2\).
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The natural logarithm has base \(e\), so converting \(\ln(x-1)=2\) to exponential form gives\[e^{2}=x-1.\nonumber\]Solving for \(x\),\[x=e^{2}+1.\nonumber\]The argument at this value is \(x-1=e^{2}>0\), so the value lies in the domain. The exact solution is \(x=e^{2}+1\), which is approximately \(8.389\).
Solve \(\log(x)+\log(x-3)=1\).
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The common logarithm has base \(10\). Condense the left side with the product law,\[\log\big(x(x-3)\big)=1,\nonumber\]then convert to exponential form,\[x(x-3)=10^{1}=10.\nonumber\]Expanding and collecting terms gives a quadratic equation,\[x^{2}-3x-10=0 \quad\Longrightarrow\quad (x-5)(x+2)=0,\nonumber\]so \(x=5\) or \(x=-2\). The original equation requires \(x>0\) and \(x-3>0\), that is, \(x>3\). The value \(x=5\) satisfies this, but \(x=-2\) makes both arguments negative and is extraneous. The solution is \(x=5\).
Solve \(\log_{5}(x+3)-\log_{5}(x-1)=1\).
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Condense the left side with the quotient law,\[\log_{5}\!\left(\dfrac{x+3}{x-1}\right)=1,\nonumber\]then convert to exponential form,\[\dfrac{x+3}{x-1}=5^{1}=5.\nonumber\]Clearing the denominator gives\[x+3=5(x-1) \quad\Longrightarrow\quad x+3=5x-5 \quad\Longrightarrow\quad 8=4x \quad\Longrightarrow\quad x=2.\nonumber\]The original equation requires \(x+3>0\) and \(x-1>0\), that is, \(x>1\). The value \(x=2\) satisfies this, so the solution is \(x=2\).
Solve \(\log_{3}(2x-1)=\log_{3}(x+4)\).
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Each side is a single logarithm with the same base, so apply the one-to-one property to equate the arguments,\[2x-1=x+4 \quad\Longrightarrow\quad x=5.\nonumber\]The original equation requires \(2x-1>0\) and \(x+4>0\), that is, \(x>\dfrac{1}{2}\). The value \(x=5\) satisfies this, so the solution is \(x=5\).
Solve \(2\log_{3}(x)=\log_{3}(4x-3)\).
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Use the power law to write the left side as a single logarithm,\[\log_{3}(x^{2})=\log_{3}(4x-3),\nonumber\]then apply the one-to-one property to equate the arguments,\[x^{2}=4x-3 \quad\Longrightarrow\quad x^{2}-4x+3=0 \quad\Longrightarrow\quad (x-1)(x-3)=0,\nonumber\]so \(x=1\) or \(x=3\). The original equation requires \(x>0\) and \(4x-3>0\), that is, \(x>\dfrac{3}{4}\). Both values exceed \(\dfrac{3}{4}\), so both are valid. The solutions are \(x=1\) and \(x=3\).
Sources
Several parts of this text use modifications from the following source:
- Wikipedia article: "Logarithm"
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