Problem 56 - Elementary and Intermediate Algebra

Question

Solve each equation. See Example \(8 .\) $$\log _{2}(x-7)+\log _{2} x=3$$

Step-by-Step Solution

Answer

The solution is \(x = 8\).

Step 1: Use Logarithmic Identity Properties

We start by using the property of logarithms that states \( ext{log} _{b} M + ext{log} _{b} N = ext{log} _{b} (MN)\). We can apply this to the given expression \( ext{log} _{2}(x-7)+ ext{log} _{2} x = ext{log} _{2}((x-7)x)\). Thus, the equation becomes \( ext{log} _{2}((x-7)x) = 3\).

Step 2: Convert the Logarithmic Equation to an Exponential Equation

Since we have \( ext{log} _{2}((x-7)x) = 3\), we can convert this to its exponential form: \((x-7)x = 2^{3}\). We know \(2^{3} = 8\), so \((x-7)x = 8\).

Step 3: Expand and Rearrange the Equation

Now, expand \((x-7)x\), to get \(x^2 - 7x = 8\). Rearrange the equation to form a standard quadratic equation: \(x^2 - 7x - 8 = 0\).

Step 4: Solve the Quadratic Equation

To solve the quadratic equation \(x^2 - 7x - 8 = 0\), we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -7,\) and \(c = -8\). Substitute these values into the formula to find \(x\).

Step 5: Substitute into the Quadratic Formula

Substitute \(a = 1\), \(b = -7\), and \(c = -8\) into the quadratic formula. Calculate the discriminant: \((-7)^2 - 4(1)(-8) = 49 + 32 = 81\). Then, find the roots: \(x = \frac{-(-7) \pm \sqrt{81}}{2(1)} = \frac{7 \pm 9}{2}\).

Step 6: Find and Check Potential Solutions

Calculate the solutions: \(x = \frac{7 + 9}{2} = 8\) and \(x = \frac{7 - 9}{2} = -1\). Ignore \(x = -1\) because it gives a negative argument for the logarithm, which is undefined. Check \(x = 8\) in the original equation: \(\text{log}_2(8-7) + \text{log}_2(8) = \text{log}_2(1) + \text{log}_2(8) = 0 + 3 = 3\). Thus, \(x = 8\) is valid.

Key Concepts

Properties of Logarithms

Logarithmic properties are essential tools that simplify complex logarithmic expressions. One of the most handy properties is the product rule, which is very useful in solving logarithmic equations. This rule states:

• \(\text{log}_b(M) + \text{log}_b(N) = \text{log}_b(MN)\)

This property allows us to combine two logarithms into a single one when they share the same base. In our original equation, \(\text{log}_2(x - 7) + \text{log}_2(x) = 3\), we use this property to combine them into \(\text{log}_2((x-7)x)\).
Another important logarithmic property frequently applied is the change of base formula, although not directly used in our problem, it's beneficial for general understanding:

• \(\text{log}_b(a) = \frac{\text{log}_k(a)}{\text{log}_k(b)}\)

These rules help solve logarithmic equations effectively by reducing complexity.

Exponential Equations

Exponential equations involve expressions where variables appear as exponents. When solving a logarithmic equation, like \(\text{log}_2((x-7)x) = 3\), we often convert it into an exponential equation to make it easier to handle.
This conversion utilizes the property of logarithms where \(\text{log}_b(M) = n\) can be rewritten in exponential form as \(M = b^n\). Applying this to our equation involves setting \((x-7)x = 2^3\).
Exposing an equation to exponential form reveals a clearer path to finding solutions. It connects logarithms to more familiar arithmetic operations and helps in transitioning to solving quadratic functions.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). Once we transform our expanded equation \((x-7)x = 8\) into \(x^2 - 7x - 8 = 0\), we employ this formula.
The quadratic formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting \(a = 1\), \(b = -7\), and \(c = -8\) from our equation gives us possible solutions. Calculating the discriminant allows us to determine:

• \(b^2 - 4ac = 49 + 32 = 81\)
• \(x = \frac{7 \pm 9}{2}\)

After calculating, we test the results within the context of the problem to ensure they fit the domain of the original logarithmic function.