Solving Logarithmic Equations

A logarithmic equation is an equation where the variable is found inside a logarithm. To solve these equations, the general strategy is to use logarithm properties to combine terms into a single logarithm, and then convert the equation into exponential form.

Key Properties to Remember

Before jumping into solving, make sure you know these fundamental logarithm rules:

• Product Rule: $\log_b(xy) = \log_b(x) + \log_b(y)$
• Quotient Rule: $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$
• Exponential Conversion: $\log_b(x) = y \iff b^y = x$
• One-to-One Property: If $\log_b(x) = \log_b(y)$, then $x = y$

The 4-Step Method

1. Condense: If there are multiple logarithms on one side of the equation, use the product or quotient rules to combine them into a single logarithm.
2. Convert: Rewrite the equation in exponential form to eliminate the logarithm. (If both sides have a single logarithm with the same base, just drop the logs using the one-to-one property).
3. Solve: Solve the resulting algebraic equation.
4. Check: This is crucial! The argument (inside) of a logarithm must always be strictly greater than zero. You must plug your answers back into the original equation to check for "extraneous" (invalid) solutions.

Example 1: Converting to Exponential Form

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

Step 1: Condense the logarithms. Use the product rule to combine the left side:

$\log_2(x(x - 2)) = 3$

Step 2: Convert to exponential form.

$2^3 = x(x - 2)$

Step 3: Solve the equation.

$8 = x^2 - 2x$

$x^2 - 2x - 8 = 0$

$(x - 4)(x + 2) = 0$

This gives us two potential solutions: $x = 4$ and $x = -2$.

Step 4: Check for extraneous solutions. If we plug $x = -2$ into the original equation, we get $\log_2(-2)$, which is undefined because we cannot take the logarithm of a negative number or zero. Therefore, $x = -2$ is an extraneous solution. The only valid solution is $x = 4$.

Example 2: Using the One-to-One Property

Solve: $\ln(x + 1) - \ln(x - 1) = \ln 3$

Step 1: Condense the logarithms. Use the quotient rule on the left side:

$\ln\left(\frac{x + 1}{x - 1}\right) = \ln 3$

Step 2: Use the one-to-one property. Since both sides are natural logs ($\ln$), we can set their arguments equal to each other:

$\frac{x + 1}{x - 1} = 3$

Step 3: Solve the equation. Multiply both sides by $(x - 1)$:

$x + 1 = 3(x - 1)$

$x + 1 = 3x - 3$

$4 = 2x$

$x = 2$

Step 4: Check your answer. Plug $x = 2$ into the original equation arguments: $(2 + 1) = 3 > 0$ and $(2 - 1) = 1 > 0$. Both arguments are positive, so $x = 2$ is a valid solution.