Question

Solve the equation. Check for extraneous solutions. \(\log _6 3 x+\log _6(x-1)=3\)

Short answer

The solution to the given equation is \(x = 9\). Note that while \(x = -8\) appeared as a mathematical solution, it's an extraneous solution as it doesn't fall within the domain of the original equation.

Step-by-step solution

Use Properties of Logarithms

Apply the property \(\log_b a + \log_b c = \log_b (a \cdot c)\) to combine the two logarithms into one: \(\log_6 (3x \cdot (x - 1)) = 3\). Which simplifies to: \(\log_6 (3x^2 - 3x) = 3\).

Convert to an Exponential Equation

Change the logarithmic equation to an exponential equation using the definition of a logarithm: if \( \log_b a = c \), then \( b^c = a \). Now the equation becomes: \(6^3 = 3x^2 -3x\). Which simplifies to: \(216 = 3x^2 -3x\).

Rearrange the Equation

Rearrange the equation into a standard quadratic form \(ax^2 + bx + c = 0\). Subtract 216 from both sides to obtain: \(3x^2 - 3x - 216 = 0\). Divide every term by 3 to simplify further: \(x^2 - x - 72 = 0\).

Solve the Quadratic Equation

Factor the quadratic equation as: \((x - 9)(x + 8) = 0\). Setting each factor equal to zero gives the potential solutions \(x = 9\) and \(x = -8\).

Check for Extraneous Solutions

We should verify these solutions don't make the argument of the log function in the original equation non-positive. Plugging in \(x = 9\) gives \(\log_6 (27) + \log_6 (8)\) both of which are positive, so \(x = 9\) is indeed a solution. However, plugging in \(x = -8\) results in \(\log_6 (-24) + \log_6 (-9)\), which are undefined as logarithm function is only defined for positive arguments. Therefore, \(x = -8\) is extraneous, and is not a solution to the original equation.

Key concepts

Properties of Logarithms

When it comes to solving logarithmic equations, understanding the properties of logarithms is crucial. Logarithms can seem daunting at first, but their properties, such as product, quotient, and power rules, make them manageable.

Combining and Simplifying Logarithmic Expressions

One of these essential properties is the ability to combine two logs with the same base into a single log by multiplying their arguments, known as the product rule, which is represented as \(\log_b a + \log_b c = \log_b (a \cdot c)\). In our example, this property was used to combine \(\log _6 3x\) and \(\log_6(x-1)\) into \(\log_6 (3x^2 - 3x)\). This step is the foundation in transforming the equation from a logarithmic form to a more solvable quadratic form.

Exponential Equations

Another key concept to consider in our exercise is the transformation of logarithmic equations into exponential equations. This is often done to take advantage of the more familiar solving methods for exponential forms.

Exponential Form of a Logarithm

Logarithmic and exponential functions are inverses of each other, and this relationship allows us to switch between the two when solving equations. The exponential form of a logarithmic equation \( \log_b a = c \) is given by \( b^c = a \) — this tells us that the base raised to the logarithm's result is equal to the logarithm's argument. For instance, we converted \(\log_6 (3x^2 - 3x) = 3\) into \(6^3 = 3x^2 -3x\), which simplifies our solving process by bringing us to a familiar quadratic equation.

Quadratic Equations

Once we have an exponential equation, our next job is to solve the resulting quadratic equation. A quadratic equation is any equation that can be rearranged in the form \( ax^2 + bx + c = 0 \).

Factoring and Solving Quadratics

Quadratic equations are staple challenges in algebra, and they can often be solved by factoring, completing the square, or using the quadratic formula. In this case, we used factoring to break down the quadratic \(x^2 - x - 72 = 0\) into two binomials \(x - 9\) and \(x + 8\), allowing us to find the roots of the equation. Solving quadratic equations effectively requires practice and a careful balance of these techniques.

Extraneous Solutions

Finally, a common pitfall in solving logarithmic equations is the emergence of extraneous solutions — solutions that mathematically solve the transformed equations but do not satisfy the original problem.

Identifying and Discarding Extraneous Solutions

For a solution to be considered valid, it must make the argument of the logarithm positive, as logs are not defined for zero or negative numbers. Hence, part of solving logarithmic equations involves verifying each potential solution. In our example, \(x = -8\) makes the log arguments negative and is therefore extraneous. Recognizing extraneous solutions is vital to ensure the final answer is correct within the domain of the original logarithmic equation.