Problem 56 - A Graphical Approach to College Algebra

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log _{2}(x-7)+\log _{2} x=3$$

Step-by-Step Solution

Answer

The solution is \( x = 8 \).

Step 1: Apply the product rule for logarithms

The given logarithmic equation is \( \log_{2}(x-7) + \log_{2} x = 3 \). Start by applying the product rule for logarithms, which states \( \log_{b} a + \log_{b} c = \log_{b}(a \cdot c) \). This gives us:\[\log_{2}((x-7) \cdot x) = 3.\]

Step 2: Exponentiate to remove the logarithm

To solve for \( x \), we need to remove the logarithm. Exponentiate both sides of the equation using base 2:\[(x-7) \cdot x = 2^3.\]Simplify the right side:\[(x-7) \cdot x = 8.\]

Step 3: Solve the quadratic equation

The equation \( x(x-7) = 8 \) simplifies to a quadratic equation. Expand and rearrange it:\[x^2 - 7x - 8 = 0.\]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -7 \), and \( c = -8 \), to find \( x \):\[x = \frac{7 \pm \sqrt{49 + 32}}{2}.\]Simplify under the square root:\[x = \frac{7 \pm \sqrt{81}}{2}.\]Further simplify, since \( \sqrt{81} = 9 \):\[x = \frac{7 \pm 9}{2}.\]This gives us the solutions \( x = 8 \) and \( x = -1 \).

Step 4: Verify the solutions

We need to verify that the solutions satisfy the original equation. Substitute \( x = 8 \):\[\log_{2}(8-7) + \log_{2}(8) = \log_{2}(1) + \log_{2}(8) = 0 + 3 = 3.\]This satisfies the original equation.For \( x = -1 \), it results in logarithms of negative numbers:\[\log_{2}(-8) + \log_{2}(-1),\]which are undefined.Therefore, \( x = -1 \) is not a valid solution. Thus, the solution \( x = 8 \) is correct.

Key Concepts

Product Rule for Logarithms

The product rule for logarithms is a handy tool for simplifying expressions with multiple logarithms. If you have an equation like \( \log_{b}(a) + \log_{b}(c) \),you can combine it using the product rule:\( \log_{b}(a) + \log_{b}(c) = \log_{b}(a \cdot c) \).
This rule is central to solving logarithmic equations efficiently.

• It allows combining the logarithmic terms.
• This results in a single term that is easier to work with.
• It is especially useful when you move on to more advanced steps like exponentiation.

In our specific problem, when we apply the product rule, \( \log_{2}(x-7) + \log_{2} x = \log_{2}((x-7) \cdot x) \),transforming two terms into one.
This step is crucial for opening the door to solving the equation with further algebraic techniques.

Quadratic Equation

When solving logarithmic equations, sometimes they transform into quadratic equations. A quadratic equation is any equation that can be rearranged in the general form:\( ax^2 + bx + c = 0 \).
To solve quadratic equations, you can use the quadratic formula:\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

• It provides solutions for any quadratic equation, provided that the determinant, \( b^2 - 4ac \), is greater than or equal to zero.
• This formula is a reliable method, especially when factoring is complex or impossible.

In our problem, after applying the product rule and simplifying, we ended up with\( x(x-7) = 8 \).By expanding and rearranging these terms, we arrive at\( x^2 - 7x - 8 = 0 \),a quadratic equation ready for the quadratic formula.
By solving this,we got the potential solutions\( x = 8 \) and \( x = -1 \).
However, it is always crucial to verify these results, as not all may satisfy the original problem fully.

Verification of Solutions

Verification is an essential step in solving equations, especially when possible solutions don't always meet the initial conditions.
By substituting each solution back into the original equation, we ensure that they hold valid within the context of the problem.

• Helps confirm whether the solutions are part of the valid domain.
• Verifies no errors were made during algebraic manipulations.
• Continues the learning process by double-checking work.

For our example, the solution \( x = 8 \) proved valid. It satisfied the equation:\( \log_{2}(8-7) + \log_{2}(8) = 3 \).
Alternately, \( x = -1 \) did not satisfy the original equation as it led to undefined logarithms.
Understanding why solutions might not work helps improve comprehension of logarithmic behavior and ensures final answers are correct and applicable.