What if there are logs on both sides of the equation?

If the equation can be written as $\log_b(A) = \log_b(B)$, use the one-to-one property to deduce that $A = B$. Equate the arguments, solve for the variable, and check for extraneous solutions.

What if the logarithms have different bases?

Use the change of base formula, $\log_b(x) = \frac{\log_c(x)}{\log_c(b)}$, to convert all logarithms to a common base before attempting to condense them.

Can a logarithmic equation have no solution?

Yes. If all algebraically derived values result in negative or zero arguments when substituted back into the original equation, they must be discarded, leaving no real solution.

How to solve a logarithmic equation

Solving a logarithmic equation requires isolating the logarithmic terms and converting the equation into exponential form. This method applies when the unknown variable is located inside the argument of one or more logarithms.

The setup

Move all terms containing logarithms to one side of the equation and constants to the other. If there are multiple logarithms with the same base, use logarithm properties (product, quotient, and power rules) to condense them into a single logarithmic expression.

The steps

Condense the logarithms into the standard form $\log_b(A) = c$ . 2. Rewrite the equation in exponential form as $A = b^c$ . 3. Solve the resulting algebraic equation (linear, quadratic, etc.) for the variable.

Checking the result

Substitute each potential solution back into the original equation. Ensure that every logarithmic argument remains strictly greater than zero. Discard any extraneous solutions that result in a zero or negative argument.

Common errors

Forgetting to check for extraneous solutions is the most frequent error. Another common mistake is improperly applying the product rule, such as writing $\log_b(x) + \log_b(y)$ as $\log_b(x+y)$ instead of the correct $\log_b(xy)$ .

Worked example

Solve for $x$ : $\log_2(x) + \log_2(x-3) = 2$

Step 1: Condense the logarithms. $\log_2(x(x-3)) = 2$ $\log_2(x^2 - 3x) = 2$

Step 2: Convert to exponential form. $x^2 - 3x = 2^2$ $x^2 - 3x = 4$

Step 3: Solve the quadratic equation. $x^2 - 3x - 4 = 0$ $(x-4)(x+1) = 0$ Potential solutions: $x = 4$ , $x = -1$

Step 4: Check for extraneous solutions. For $x = 4$ : $\log_2(4) + \log_2(4-3) = 2 + 0 = 2$ . (Valid) For $x = -1$ : $\log_2(-1)$ is undefined. (Extraneous)

The only solution is $x = 4$ .

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References: OpenStax College Algebra, Chapter 6: Exponential and Logarithmic Equations · Stewart Algebra and Trigonometry, Chapter 4: Exponential and Logarithmic Functions