Problem 62 - College Algebra

Question

For the following exercises, solve the equation for \(x\), if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution. $$ \log \left(x^{2}+13\right)=\log (7 x+3) $$

Step-by-Step Solution

Answer

The solutions to the equation are \( x = 5 \) and \( x = 2 \).

Step 1: Understand the equation

The given equation is \( \log(x^2 + 13) = \log(7x + 3) \). We need to solve for \( x \). Realize that if the logarithms are equal, their arguments must be equal.

Step 2: Set the arguments equal

Since \( \log(a) = \log(b) \) implies \( a = b \), we can set the arguments equal: \( x^2 + 13 = 7x + 3 \).

Step 3: Rearrange the equation

Rearrange the equation by moving all terms to one side: \( x^2 - 7x + 13 - 3 = 0 \). Simplify it to get \( x^2 - 7x + 10 = 0 \).

Step 4: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -7 \), and \( c = 10 \). First, compute the discriminant: \( b^2 - 4ac = (-7)^2 - 4 \cdot 1 \cdot 10 = 49 - 40 = 9 \).

Step 5: Calculate the roots

Now calculate the roots: \( x = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2} \). This gives the solutions \( x = 5 \) and \( x = 2 \).

Step 6: Verify solutions in the original equation

Substitute \( x = 5 \) and \( x = 2 \) back into the original arguments to ensure they are valid: - For \( x = 5 \): \( \log(5^2 + 13) = \log(38) \) and \( \log(7 \cdot 5 + 3) = \log(38) \), so \( x = 5 \) is valid.- For \( x = 2 \): \( \log(2^2 + 13) = \log(17) \) and \( \log(7 \cdot 2 + 3) = \log(17) \), so \( x = 2 \) is valid.

Step 7: Graph both sides

Graph the functions \( y = \log(x^2 + 13) \) and \( y = \log(7x + 3) \). Identify the points where the graphs intersect. The points of intersection correspond to the solutions \( x = 5 \) and \( x = 2 \).

Key Concepts

Quadratic Formula

The quadratic formula is a tool used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). To find the values of \( x \) that satisfy the equation, we use the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The term under the square root, \( b^2 - 4ac \), is known as the discriminant and it determines the nature of the roots:

• If the discriminant is positive, the equation has two real and distinct roots.
• If it is zero, there are exactly two real and identical roots (also called a repeated root).
• If negative, the roots are complex or imaginary, meaning they have no real number solutions.

In our exercise, we applied the quadratic formula to solve the equation \( x^2 - 7x + 10 = 0 \), leading to solutions \( x = 5 \) and \( x = 2 \). Understanding how the discriminant affects the solutions is crucial to predicting the nature of the solutions before even computing them.

Graphing Functions

Graphing functions provides a visual understanding of equations, illustrating how different values influence the terms. In our exercise, we graphed the functions \( y = \log(x^2 + 13) \) and \( y = \log(7x + 3) \). By plotting these equations, we could visually inspect where both sides of the original equation are equal, confirming our algebraic solutions.

• The logarithmic nature of the functions means they only accept positive arguments, since the logarithm of zero or a negative number is undefined.
• It's vital to consider this when solving any logarithmic equation, as invalid real-world solutions would surface visibly when graphed.

Graphing offers an intuitive check of your solutions, demonstrating symmetry, differences, and confirming the validity of our algebraically found solutions.

Point of Intersection

The point of intersection between two graphs is where their plotted functions reach the same y-value for the same x-value. In context, these points are solutions to the equation being studied. Through graphing, it becomes possible to visualize these intersections, which correspond to solutions we've derived algebraically. For the given exercise, we found two points of intersection on the graphs of \( y = \log(x^2 + 13) \) and \( y = \log(7x + 3) \), at \( x = 5 \) and \( x = 2 \). At these points, both functions have the same values, supporting the solutions we calculated using algebraic methods.Visual confirmation helps ensure both the correctness and understanding of solutions. It serves as a bridge between numerical values and geometric representation.

Validating Solutions

Validating solutions involves verifying that the solutions calculated meet all the conditions of the original problem. For logarithmic equations, this means that any proposed solution, when substituted back, should make the arguments of all logarithms involved positive numbers. In our exercise, after solving, we found \( x = 5 \) and \( x = 2 \), and validated by substituting back into the equation:

• For \( x = 5 \): Substituting into \( \log(5^2 + 13) = \log(38) \) and \( \log(7 \cdot 5 + 3) = \log(38) \), we see both sides match, confirming \( x = 5 \) is indeed a valid solution.
• For \( x = 2 \): Substituting into \( \log(2^2 + 13) = \log(17) \) and \( \log(7 \cdot 2 + 3) = \log(17) \), both again match, confirming \( x = 2 \) as valid.

Validating solutions ensures they adhere to any constraints imposed by the equation, confirming they're mathematically and practically acceptable.