Problem 80 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log _{5}(7+x)+\log _{5}(8-x)-\log _{5} 2=2 $$

Step-by-Step Solution

Answer

The equation has no real solutions, only complex solutions.

Step 1: Reformulate the Logarithmic Equation

First, recognize that we can use the properties of logarithms to combine the terms on the left-hand side of the equation. Use the product and quotient rules of logarithms: \[ \log_b M + \log_b N = \log_b (MN) \] and \[ \log_b M - \log_b N = \log_b \left( \frac{M}{N} \right) \].Applying these properties, we combine:\[ \log_5((7+x)(8-x)) - \log_5 2 = 2 \]This further simplifies to:\[ \log_5 \left(\frac{(7+x)(8-x)}{2}\right) = 2 \]

Step 2: Remove Logarithms and Solve Quadratic

To eliminate the logarithm, rewrite the equation in exponential form:\[ \frac{(7+x)(8-x)}{2} = 5^2 \]This simplifies to:\[ (7+x)(8-x) = 50 \times 2 = 100 \]Expand the product on the left-hand side:\[ 56 - x + 7x - x^2 = 100 \]Combine like terms to form a quadratic equation:\[ -x^2 + 6x + 56 = 100 \]Rearrange into standard quadratic form:\[ x^2 - 6x + 44 = 0 \]

Step 3: Solve the Quadratic Equation

Use the quadratic formula to find the roots of the quadratic equation:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For the equation \(x^2 - 6x + 44 = 0\), the coefficients are: \(a = 1\), \(b = -6\), and \(c = 44\).Substitute these values into the quadratic formula:\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 44}}{2 \times 1} \]\[ x = \frac{6 \pm \sqrt{36 - 176}}{2} \]\[ x = \frac{6 \pm \sqrt{-140}}{2} \]Since the discriminant is negative, there are no real solutions, and the solutions are complex. Thus, we cannot provide real solutions as requested in the problem.

Step 4: Conclusion

Because the discriminant is negative, the original equation has no real solutions. The solutions, if considered, are complex and not necessary for an approximation to four decimal places.

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations are called 'quadratic' because the highest power of the variable \(x\) is 2. Quadratics can be solved by different methods including factoring, completing the square, or using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The discriminant, \(b^2 - 4ac\), is very important in determining the nature of the solutions:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is one real root, also known as a repeated or double root.
• If the discriminant is negative, the roots are complex numbers and not real.

In solving the original problem, we used the quadratic formula and found a negative discriminant, indicating complex solutions.

Properties of Logarithms

Logarithms have several key properties that help transform and simplify logarithmic equations. These include:

• The product property: \(\log_b(MN) = \log_b M + \log_b N\)
• The quotient property: \(\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N\)
• The power property: \(\log_b(M^p) = p \cdot \log_b M\)

These properties are used to combine multiple logarithmic expressions into a single logarithm or to break down complex logarithms. In the given problem, the product and quotient properties were used to combine logarithmic terms, leading to a simpler form. This simplification allowed us to transform the logarithmic equation into a quadratic form that can be handled using algebraic methods.

Complex Numbers

Complex numbers emerge when the discriminant of a quadratic equation is negative, leading to solutions that are not on the real number line. A complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).
In algebra, complex numbers provide solutions to equations that have no real solutions.
When we solved the quadratic equation \(x^2 - 6x + 44 = 0\), the negative discriminant \(-140\) resulted in solutions involving complex numbers:
\[x = \frac{6 \pm \sqrt{-140}}{2}\]
By introducing complex numbers, we can extend our understanding of solutions to include those that exist in a plane of both real and imaginary parts.