Problem 28 - Algebra A Combined Function

Question

Solve each equation. $$ \log _{8}\left(x^{2}-2 x\right)=1 $$

Step-by-Step Solution

Answer

The solutions are \( x = 4 \) and \( x = -2 \).

Step 1: Understand the Given Equation

We have the equation \( \log_{8}(x^2 - 2x) = 1 \). This is a logarithmic equation where the base is 8.

Step 2: Convert Logarithmic to Exponential Form

Recall that \( \log_b(a) = c \) is equivalent to \( b^c = a \). Here, \( b = 8 \), \( a = x^2 - 2x \), and \( c = 1 \). Thus, we convert this to exponential form: \( 8^1 = x^2 - 2x \).

Step 3: Simplify the Exponential Equation

The equation \( 8^1 = x^2 - 2x \) simplifies to \( 8 = x^2 - 2x \). Rearranging, we get \( x^2 - 2x - 8 = 0 \).

Step 4: Solve the Quadratic Equation

The quadratic equation \( x^2 - 2x - 8 = 0 \) can be factored as \( (x - 4)(x + 2) = 0 \). Set each factor equal to zero: \( x - 4 = 0 \) or \( x + 2 = 0 \).

Step 5: Find the Solutions

Solving these equations gives \( x = 4 \) and \( x = -2 \). However, we must verify if these solutions satisfy the original log equation.

Step 6: Verify the Solutions

Substitute \( x = 4 \) into the original equation: \( x^2 - 2x = 4^2 - 2 \times 4 = 16 - 8 = 8 \). This satisfies the equation. Now, substitute \( x = -2 \): \( (-2)^2 - 2 \times (-2) = 4 + 4 = 8 \). This also satisfies the equation.

Key Concepts

Exponential Form

When solving logarithmic equations, it's useful to understand how to switch between logarithmic and exponential forms since each can make different types of problems easier to handle. In a logarithmic expression like \( \log_b(a) = c \), "b" represents the base of the logarithm, "a" is the result, and "c" is the exponent. To convert this into exponential form, we express it as \( b^c = a \).

For the exercise \( \log_8(x^2 - 2x) = 1 \), our task is to express this equation exponentially. Applying the conversion, we get \( 8^1 = x^2 - 2x \), which simplifies to \( 8 = x^2 - 2x \). This step is crucial because it results in an equation form that we can further manipulate using algebraic techniques.

Understanding these conversions can help solve equations more efficiently and recognize that logarithmic expressions can often be seen as another representation of their exponential counterparts.

Quadratic Equations

Once the logarithmic equation has been converted to exponential form (\( 8 = x^2 - 2x \)), the next step is often to solve a quadratic equation. A quadratic equation is typically represented as \( ax^2 + bx + c = 0 \). In our case, rearranging the terms gives us \( x^2 - 2x - 8 = 0 \).

Quadratic equations can be solved using various methods such as factoring, completing the square, or using the quadratic formula. Here, the equation \( x^2 - 2x - 8 = 0 \) can be easily factored to \( (x - 4)(x + 2) = 0 \).

By setting each factor equal to zero, we find the potential solutions:

• \( x - 4 = 0 \) gives \( x = 4 \).
• \( x + 2 = 0 \) gives \( x = -2 \).

These solutions are proposed answers but must be verified as valid solutions to the original equation.

Equation Verification

Verification involves ensuring that the solutions found truly satisfy the original equation. This is an essential step because operations performed during the solution process might introduce extraneous solutions that don't actually fit the initial equation.

To verify the solutions in \( \log_8(x^2 - 2x) = 1 \), substitute each solution back into the original expression. Starting with \( x = 4 \), we substitute into \( x^2 - 2x \) which results in:

• \( 4^2 - 2 \times 4 = 16 - 8 = 8 \)

This result 8, when re-applied to the logarithmic function, should equate to the power of 1 since \( \log_8(8) = 1 \).

Next, for \( x = -2 \):

• \((-2)^2 - 2 \times (-2) = 4 + 4 = 8\)

Again, this confirms \( \log_8(8) = 1 \).

Both substitutions show that the solutions are correct, and effectively solving any equation requires this crucial verification step to avoid invalid or extraneous results.