Problem 42 - College Algebra

Question

For the following exercises. use the one-to-one pronerty of logarithms to solve. $$ \log _{9}\left(2 n^{2}-14 n\right)=\log _{9}\left(-45+n^{2}\right) $$

Step-by-Step Solution

Answer

The solutions are \( n = 5 \) and \( n = 9 \).

Step 1: Apply the One-to-One Property

The one-to-one property states that if \( \log_b(x) = \log_b(y) \) for any logarithm with the same base, \( x = y \). Here, both expressions have a logarithm base of 9, so we can equate the arguments: \( 2n^2 - 14n = -45 + n^2 \).

Step 2: Simplify the Equation

Rearrange the equation by moving all terms to one side: \( 2n^2 - 14n - n^2 + 45 = 0 \). Simplifying gives \( n^2 - 14n + 45 = 0 \).

Step 3: Factor the Quadratic Equation

The equation \( n^2 - 14n + 45 = 0 \) is a quadratic that can be factored. We look for two numbers that multiply to 45 and add up to -14. These numbers are -5 and -9: \((n-5)(n-9) = 0\).

Step 4: Solve for n

Set each factor equal to zero: \( n-5 = 0 \) and \( n-9 = 0 \). Solving these gives \( n = 5 \) and \( n = 9 \).

Step 5: Verify the Solutions

Substitute \( n = 5 \) and \( n = 9 \) back into the original expressions to ensure they satisfy the equation. Both \( n = 5 \) and \( n = 9 \) work, as they make both sides of the logarithmic equation equal. Therefore, both are valid solutions.

Key Concepts

Logarithmic Equations

Logarithmic equations are equations that involve logarithms with unknown variables. To solve these equations effectively, the one-to-one property of logarithms comes in handy. This property allows us to equate the arguments of two logarithms when they are equal and share the same base. For example, if you have two logarithmic expressions like

• \( \log_b(x) = \log_b(y) \)

Then it follows that

• \( x = y \)

This elimination of the logarithm through the one-to-one property simplifies the process of solving for the variable within the logarithmic equation.
Once the equation is simplified to involve a basic polynomial, you can proceed with algebraic methods, such as solving quadratics or factoring, to find the unknown variable.

Solving Quadratics

Quadratic equations are polynomial equations of the form

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable.
In our specific case of solving quadratics that arise from logarithmic equations, we first simplify the logarithmic equation into a quadratic form, just like converting

• \( 2n^2 - 14n = -45 + n^2 \)

into

• \( n^2 - 14n + 45 = 0 \)

With the quadratic equation formed, we have multiple methods to solve it:

• The quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{2a} \)
• Completing the square
• Factoring if simple enough

In cases where factoring is straightforward, it saves time and effort compared to other methods. The key to solving these equations is accurately forming and simplifying them first.

Factoring Quadratics

Factoring quadratics is a common technique used to simplify and solve quadratic equations.
This involves expressing the quadratic as a product of two binomial expressions. For example, a quadratic like

• \( n^2 - 14n + 45 = 0 \)

can be rewritten as

• \((n-5)(n-9) = 0\)

The factorization requires finding two numbers that multiply to the constant term \( c \) (in this instance, 45) and add up to the linear coefficient \( b \) (here, -14).
Once factored, solving the quadratic is as simple as setting each of the expressions equal to zero:

• \( n - 5 = 0 \)
• \( n - 9 = 0 \)

These equations solve to give \( n = 5 \) and \( n = 9 \) respectively. Thus, factoring not only simplifies the equation but also leads directly to its solutions.