We have already seen that every logarithmic equation log_b(x) = y is equivalent to the exponential equation b^y = x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. For example, consider the equation log ₂(2) + log₂(3x – 5) = 3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x:

	[latex]\displaystyle{{log}_{{2}}{({2})}}+{{log}_{{2}}{({3}{x}-{5})}}={3}[/latex]
Apply the product rule of logarithms	[latex]\displaystyle{{log}_{{2}}{({2}{({3}{x}-{5})})}}={3}[/latex]
Distribute	[latex]\displaystyle{{log}_{{2}}{({6}{x}-{10})}}={3}[/latex]
Apply the definition of a logarithm	[latex]\displaystyle{2}^{{3}}={6}{x}-{10}[/latex]
Calculate 2 3	[latex]\displaystyle{8}={6}{x}-{10}[/latex]
Add 10 to both sides	[latex]\displaystyle{18}={6}{x}[/latex]
Divide by 6	[latex]\displaystyle{x}={3}[/latex]

Using the Definition of a Logarithm to Solve Logarithmic Equations

For any algebraic expression S and real numbers b and c, where b > 0, b ≠ 1, log _b(S) = c if and only if b^c = S

Example 1

Using algebra to solve a logarithmic equation

1. Solve [latex]\displaystyle{2}{ln{{x}}}+{3}={7}[/latex]
   1. Subtract 3: [latex]\displaystyle{2}{ln{{x}}}={4}[/latex]
   2. Divide by 2: [latex]\displaystyle{ln{{x}}}={2}[/latex]
   3. Rewrite in exponential form: [latex]\displaystyle{x}={e}^{{2}}[/latex]
2. Solve [latex]\displaystyle{6}+{ln{{x}}}={10}[/latex]

Using algebra before and after using the definition of the natural logarithm

1. Solve [latex]\displaystyle{2}{ln{{({6}{x})}}}={7}[/latex]
   1. Divide by 2: [latex]\displaystyle{ln{{({6}{x})}}}=\frac{{7}}{{2}}[/latex]
   2. Use the definition of ln: [latex]\displaystyle{6}{x}={e}^{{{(\frac{{7}}{{2}})}}}[/latex]
   3. Divide by 6: [latex]\displaystyle{x}=\frac{{1}}{{6}}{e}^{{{(\frac{{7}}{{2}})}}}[/latex]
2. Solve [latex]\displaystyle{2}{ln{{({x}+{1})}}}={10}[/latex]
3. Using a graph to understand the solution to a logarithmic equation, solve [latex]\displaystyle{ln{{x}}}={3}[/latex]The graphs of y = lnxand y = 3 cross at the point (e³,3) which is approximately (20.0855, 3).
4. Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2^x = 1000, solve to 2 decimal places. x ≈ 9.97

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where b ≠ 1, log_bS = log_bT if and only if S = T For example, If log₂(x − 1) = log₂(8), then x − 1 = 8. So, if x – 1 =8, then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: log₂(9 – 1) = log₂(8) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown. For example, consider the equation [latex]\displaystyle{log{{({3}{x}-{2})}}}-{log{{({2})}}}={log{{({x}+{4})}}}[/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for [latex-display]\displaystyle{log{{({3}{x}-{2})}}}-{log{{({2})}}}={log{{({x}+{4})}}}[/latex-display]

Apply the quotient rule of logarithms	[latex]\displaystyle{log{{(\frac{{{3}{x}-{2}}}{{2}})}}}={log{{({x}+{4})}}}[/latex]
Apply the one to one property of a logarithm.	[latex]\displaystyle\frac{{{3}{x}-{2}}}{{2}}={x}+{4}[/latex]
Multiply both sides of the equation by 2	[latex]\displaystyle{3}{x}-{2}={3}{x}+{8}[/latex]
Subtract 2 x and add 2	[latex]\displaystyle{x}={10}[/latex]

To check the result, substitute into [latex-display]\displaystyle{log{{({3}{({10})}-{2})}}}-{log{{({2})}}}={log{{({({10})}+{4})}}}[/latex-display] [latex-display]\displaystyle{log{{({28})}}}-{log{{({2})}}}={log{{({14})}}}[/latex-display] [latex-display]\displaystyle{log{{(\frac{{28}}{{2}})}}}={log{{({14})}}}[/latex-display] The solution checks.

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

For any algebraic expressions S and T and any positive real number b, where b ≠ 1, log_bS = log_bT if and only if S = T Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution. Given an equation containing logarithms, solve it using the one-to-one property.

1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form log_bS = log_bT.
2. Use the one-to-one property to set the arguments equal.
3. Solve the resulting equation, S = T, for the unknown.

Example 2

1. Using the one-to-one property of logarithms, solve [latex]\displaystyle{ln{{({x}^{{2}})}}}={ln{{({2}{x}+{3})}}}[/latex]Use the one-to one property of the logarithm: [latex]\displaystyle{x}^{{2}}={2}{x}+{3}[/latex]Get zero on one side before factoring: [latex]\displaystyle{x}^{{2}}-{2}{x}-{3}={0}[/latex]Factor using FOIL: [latex]\displaystyle{({x}-{3})}{({x}+{1})}={0}[/latex] If a product is zero, one of the factors must be zero: [latex]\displaystyle{x}-{3}={0}{\text{ or }}{x}+{1}={0}[/latex] Solve for x: [latex]\displaystyle{x}={3}{\text{ or }}{x}=-{1}[/latex] Analysis There are two solutions: x = 3 or x = –1. The solution x = –1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.
2. Solve [latex]\displaystyle{ln{{({x}^{{2}})}}}={ln{{1}}}[/latex]

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OpenStax, Precalculus, "Exponential and Logarithmic Equations," licensed under a CC BY 3.0 license.