Problem 57 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{6} x=1-\log _{6}(x-5)$$

Step-by-Step Solution

Answer

Solution: 1. Set the logarithms equal to each other: \(\log_6 x + \log_6 (x-5) = 1\). 2. Use product rule: \(\log_6 (x(x-5)) = 1\). 3. Remove logarithm: \(6^1 = x(x-5)\) and simplify: \(6 = x^2 - 5x\). 4. Set equation to standard form: \(x^2 - 5x - 6 = 0\). 5. Factor the quadratic equation: \((x - 6)(x + 1) = 0\). 6. Solve for x: \(x = 6\) and \(x = -1\). 7. Check validity: \(x = 6\) is valid, \(x = -1\) is not valid. 8. Check with graphing calculator: intersection point is at \(x = 6\). The valid algebraic solution is \(x = 6\).

Step 1: Set the logarithms equal to each other

From the given equation, we want to set both logarithms equal to each other, so we can add \(\log _6(x-5)\) to both sides: \[\log_6 x + \log_6 (x-5) = 1.\]

Step 2: Use properties of logarithm

Now, we can use the properties of logarithms to simplify the equation. In this case, we'll use the product rule which states \(\log_b a + \log_b c = \log_b (a * c)\). Applying this to our equation, we get: \[\log_6 (x(x-5)) = 1.\]

Step 3: Simplify the equation

Now we want to simplify the equation by removing the logarithm. We can do this by using the definition of logarithm \(\log_b a = c \Leftrightarrow b^c = a\): \[6^1 = x(x-5).\] Simplify: \[6 = x^2 - 5x.\]

Step 4: Rearrange the equation

Now we need to solve for x which involves solving a quadratic equation. First, set the equation to standard form by moving all terms to the left side: \[x^2 - 5x - 6 = 0.\]

Step 5: Factor the quadratic equation

Factor the quadratic equation to find the possible values of x: \[(x - 6)(x + 1) = 0.\]

Step 6: Solve for x

Now, we have two possible solutions for x: \[x - 6 = 0 \Rightarrow x = 6,\] \[x + 1 = 0 \Rightarrow x = -1.\]

Step 7: Check validity of solutions

We need to make sure both solutions are valid by seeing if they satisfy the original equation. In logarithmic functions, we cannot take a log of a negative number or zero, so we can discard the solution x = -1 since it makes one of the logarithm arguments negative. Therefore, the valid algebraic solution is \(x = 6\).

Step 8: Check with a graphing calculator

Now, check the solution using a graphing calculator. Plot the two original functions: \(f(x) = \log _6 x, g(x) = 1 - \log _6(x-5)\), and find their intersection point. The intersection point will have the same x-coordinate as our valid solution above. Using the graphing calculator, we can confirm that the intersection point is at \(x = 6\), which matches our algebraic solution.

Key Concepts

Quadratic Equation

Quadratic equations are polynomial equations of degree two. They take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solving quadratic equations involves finding the values of \(x\) that make the equation true. In our solution, we arrived at a quadratic equation \(x^2 - 5x - 6 = 0\).

To solve it, we factor it into simpler expressions: \((x - 6)(x + 1) = 0\). This gives us the possible solutions \(x = 6\) and \(x = -1\). Quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In this context, the solution \(x = 6\) is valid, as it satisfies the conditions for the original logarithmic equation, while \(x = -1\) does not.

Properties of Logarithms

Logarithms have several important properties that simplify solving equations. In this example, we use the **product rule**, which states \(\log_b a + \log_b c = \log_b (a \cdot c)\). This property allows us to combine two logarithms into a single logarithm.

For our specific exercise, we utilized this property to combine \(\log_6 x + \log_6 (x-5)\) into \(\log_6 (x(x-5))\). Understanding these properties is crucial in manipulating and simplifying logarithmic expressions during problem-solving.

Additionally, when simplifying logarithmic equations, recognizing the definition \(\log_b a = c\) translates to \(b^c = a\) is also crucial. It lets you rewrite complex logarithmic equations into more manageable exponential forms.

Algebraic Solution

Solving logarithmic equations algebraically involves using algebraic manipulations to isolate the variable. This often includes applying properties of logarithms, as well as converting logs into exponential form as needed.

For the equation \(\log_6 x = 1 - \log_6(x-5)\), we first set the terms with logarithms equal and then apply log properties to combine them. This led us to \(\log_6 (x(x-5)) = 1\) which simplifies to \(6^1 = x(x-5)\), eventually becoming \(x^2 - 5x - 6 = 0\).

Converting it into a quadratic equation enabled us to use techniques like factoring to find solutions for \(x\). Verifying these solutions within the context of the original problem is a key step, especially in logarithm problems, because logarithmic functions have domain restrictions.

Graphical Solution

The graphical solution involves using a graph to visually identify solutions to the equation. In this case, we use the original functions: \(f(x) = \log_6 x\) and \(g(x) = 1 - \log_6 (x-5)\).

By plotting these on a graph, we look for the intersection point, which gives the solution to the equation. Here, the intersection point confirmed our algebraic solution \(x = 6\).

Graphing can be especially helpful to check the validity of solutions, as it provides a visual interpretation of where the values satisfy the equation. It also illustrates any restrictions in the domain that we must be mindful of, such as avoiding negative or zero arguments in logarithms.