Logarithmic equations are equations in which the unknown appears inside a logarithm. To solve them, it is crucial to understand the properties of logarithms and how these can be applied to isolate and determine the value of the unknown. A logarithmic equation takes the form:

\[ \log_af(x) = g(x) \]

  • \( a \) is the base of the logarithm and and it must meet the condition \( a \gt 0, a\neq 1 \)

  • \(f(x)\), the argument of the logarithm must be greater than zero. This is because the logarithm function is only defined for positive numbers.

Solving logarithmic equations requires a deep understanding of logarithmic properties. These properties, such as the product, quotient, and power rules, are essential for simplifying and reducing the original equation to a more manageable form. Knowledge of these rules allows one to transform complex logarithmic expressions into simpler ones, making it easier to isolate the variable and find the solution.

Resolution Method

The solution method for logarithmic equations can be broken down into four fundamental steps:

  • Determine the set of admissible solutions by ensuring the arguments of the logarithms in the equation are greater than 0 and remembering that the base must be greater than 0 and different from 1. \[ \log_af(x) = \begin{cases} a > 0 \\[0.6em] a \neq 1 \\[0.6em] f(x) > 0 \\ \end{cases} \]

  • Apply logarithmic properties, such as the product, quotient, and power rules, to combine and simplify the logarithmic terms. This step aims to reduce the initial equation to a simpler form that allows for isolating the variable.

  • Once the equation is simplified, solve for the variable by eliminating the logarithms. This often involves exponentiating both sides of the equation to remove the logarithmic functions.

  • Substitute the solutions back into the original equation to ensure that they satisfy the equation. Check that all solutions are within the domain of the original logarithmic functions.

Once a logarithmic equation is simplified to a form suitable for solving the variable, the following strategies may occur:

  • The equation is reduced to the form: \[ \log_a f(x) = \log_a g(x) \] In this case, we can use the fact that when two logarithms with the same base are equal, their arguments must also be equal. Therefore, it is sufficient to set the two arguments equal and solve the resulting equation for the variable \( x \): \[ f(x) = g(x) \]

  • The equation is reduced to the form: \[ \log_a f(x) = b\] In this case, since no further simplifications can be made, the solution requires using the definition of the logarithm and converting to the exponential form, resulting in: \[ x = a^b \]

  • The equation is reduced to the form \[\log_a^2(x+c) + \log_a(x+c) + k = 0\] where logarithms raised to a certain power \(n\) appear, with the same base and the same argument. In this case, we can use a substitution of the type \(z = \log_a(x+c)\), obtaining: \(z^2 + z + k\). The equation is thus transformed into a quadratic equation or an equation of degree \(n\), which can be solved using the quadratic formula or Ruffini’s rule for polynomials of degree \(n > 2\).


Let’s solve the logarithmic equation:

\[ \log_3(2x + 1) = \log_3(x^2) \]

The first step is to determine the domain of valid solutions, ensuring that the arguments of the logarithms are positive. Therefore, we set the conditions:

\[ 2x + 1 > 0 \implies x > -\frac{1}{2} \] \[ x^2 > 0 \implies x \neq 0 \]

Since the logarithms have the same base and are in the form \( \log_a f(x) = \log_a g(x) \), we can equate the arguments directly. Rearranging the terms to one side to form a standard quadratic equation, we get:

\[ x^2-2x-1 = 0 \]

We can solve this using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]

where \( a = 1 \), \( b = -2 \), and \( c = -1 \). Substituting these values, we get:

\begin{align*} x &= \frac{-(-2) \pm \sqrt{(-2)^2-4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \\[0.6em] &= \frac{2 \pm \sqrt{4 + 4}}{2} \\[0.6em] &= \frac{2 \pm \sqrt{8}}{2} \\[0.6em] &= \frac{2 \pm 2\sqrt{2}}{2} \\[0.6em] &= 1 \pm \sqrt{2} \end{align*}

Now let’s verify that the solutions satisfy the domain:

  • \( x = 1 + \sqrt{2} \) satisfies \( x > -\frac{1}{2} \).
  • \( x = 1-\sqrt{2} \) does not satisfy \( x > -\frac{1}{2} \).

The solution to the equation is:

\[ x = 1 + \sqrt{2} \]

The above is just a basic introductory example. Please explore the dedicated exercise section to practice with problems covering all the abovementioned cases.