Introduction
A logarithmic equation can often be solved by using the relationship between logarithms and exponentials. The main step is to rewrite the logarithmic statement in exponential form, solve the resulting equation, and make sure the answer is valid in the original equation.
This tutorial focuses on equations of the form \(\log_b(g(x))=c\), where \(b\) is a fixed base, \(g(x)\) is an expression involving the unknown \(x\), and \(c\) is a fixed real number. The method applies to bases greater than \(1\) and to bases between \(0\) and \(1\), provided the base is positive and not equal to \(1\).
What Makes a Logarithmic Equation Defined?
For \(\log_b(u)\) to be defined as a real number, the base must satisfy \(b>0\) and \(b\ne1\), and the argument must satisfy \(u>0\). The argument is the expression to which the logarithm is applied.
Therefore, an equation \(\log_b(g(x))=c\) is defined only for values of \(x\) for which \(g(x)>0\). The base \(b\) must also meet the conditions above. These requirements are part of solving the equation, not optional checks to perform only if something looks unusual.
In the earlier material on converting between logarithmic and exponential form, the defining relationship is expressed as \(\log_b(u)=c\) exactly when \(b^c=u\). Applying that relationship gives the central method here.
Why the Conversion Works
The conversion is justified by the definition of the logarithm. If \(\log_b(g(x))=c\), the definition says that \(c\) is the exponent to which \(b\) must be raised to obtain \(g(x)\). Hence \(g(x)=b^c\). Conversely, if \(g(x)=b^c\), then \(b^c=g(x)\), so the exponent that produces \(g(x)\) from base \(b\) is \(c\). Because \(b^c>0\), the logarithm is defined and \(\log_b(g(x))=c\).
This equivalence assumes \(b>0\), \(b\ne1\), and a real value of \(c\). The original logarithmic equation also requires \(g(x)>0\). The converted equation enforces that condition automatically when it is solved exactly, because \(b^c\) is positive. Still, checking proposed answers in the original equation is a good way to catch arithmetic errors or mistakes made while solving the converted equation.
A Reliable Solving Process
The base must be positive and different from \(1\). The argument must be positive for any proposed value of the unknown.
Replace \(\log_b(g(x))=c\) with \(g(x)=b^c\).
Use appropriate algebra to find candidate values of \(x\).
Check that the logarithm’s argument is positive and substitute each candidate into the original equation.
Worked Examples
Worked Example: Solve a Linear Logarithmic Equation
Solve \(\log_3(2x-1)=2\).
The base is \(3\), which is positive and not equal to \(1\). The argument must satisfy \(2x-1>0\), or \(x>\frac12\). Convert the equation to exponential form:
Evaluate the power and solve:
Check the argument at this value: \(2(5)-1=9>0\), so the logarithm is defined. Substitute into the original equation:
Since \(3^2=9\), the logarithm equals \(2\). Thus \(x=5\) is the solution.
Worked Example: Solve a Logarithmic Equation That Produces Two Solutions
Solve \(\log_2(x^2-1)=3\).
The base \(2\) is valid. The logarithm requires \(x^2-1>0\). Convert to exponential form:
Evaluate the power and solve the resulting quadratic equation:
Check both values in the argument. For \(x=3\), \(x^2-1=9-1=8>0\). For \(x=-3\), \(x^2-1=(-3)^2-1=9-1=8>0\). Both logarithms are defined. Now substitute into the original equation:
Both values satisfy the equation, so the solutions are \(x=3\) and \(x=-3\).
Worked Example: A Zero on the Right-Hand Side
Solve \(\log_5(7-2x)=0\).
The base \(5\) is valid. The argument condition is \(7-2x>0\), which gives \(x<\frac72\). Convert to exponential form. Remember that any valid base raised to the power \(0\) equals \(1\):
Now solve:
The candidate meets the argument condition because \(3<\frac72\). Substitution verifies the original equation:
The last equality holds because \(5^0=1\). Therefore \(x=3\) is the solution.
Worked Example: A Base Between Zero and One
Solve \(\log_{\frac12}(x+4)=-2\).
The base \(\frac12\) is positive and not equal to \(1\), so it is a valid logarithm base. The argument must be positive: \(x+4>0\), or \(x>-4\). Convert to exponential form:
A negative exponent gives the reciprocal raised to the corresponding positive power:
The argument at \(x=0\) is \(0+4=4>0\). Check in the original equation: \(\left(\frac12\right)^{-2}=4\), so \(\log_{\frac12}(4)=-2\). Thus \(x=0\) is the solution. This example shows why a base need not be greater than \(1\); it only needs to be positive and different from \(1\).
A Common Conversion Mistake
A frequent error is to use the number \(c\) as the base when converting, instead of using the actual logarithm base \(b\). For example, consider \(\log_3(x+1)=2\). A mistaken conversion might be:
This reverses the roles of the base and the exponent. Substituting \(x=7\) into the original equation gives \(\log_3(8)\), which is not \(2\), because \(3^2=9\), not \(8\). The correct conversion keeps the logarithm’s base as the exponential base and the right-hand side as the exponent:
Checking confirms \(\log_3(8+1)=\log_3(9)=2\). A useful way to avoid this error is to say aloud: “The base raised to the right-hand side equals the argument.”
Why the Check Matters
The conversion itself preserves the equation when the logarithm is defined. Errors can nevertheless enter during algebra: a power may be evaluated incorrectly, a quadratic may be solved incorrectly, or a condition on the argument may be overlooked. Checking each candidate in the original equation catches these problems.
For an equation of the form \(\log_b(g(x))=c\), a successful check has two parts. First, confirm \(g(x)>0\), so the logarithm exists as a real number. Second, confirm that the resulting logarithm has value \(c\), equivalently that \(b^c=g(x)\). When an equation produces more than one candidate, check each one separately.
Check Your Understanding
For each question, use exponential form and check the proposed solution in the original equation.
- Solve \(\log_4(x-2)=2\).
- Solve \(\log_3(x^2)=2\). How many real solutions are there?
- Solve \(\log_2(5-x)=1\).
- In \(\log_7(2x+1)=3\), which number is the base and which number is the exponent after conversion?
- Why must the argument \(g(x)\) be positive in \(\log_b(g(x))=c\)?
