Problem 93 - Intermediate Algebra

Question

Look Alikes \(\cdots\) a. \(\log _{2}(x+5)-\log _{2} 4 x=\log _{2} x\) b. \(\ln (x+5)-\ln 4 x=\ln x\)

Step-by-Step Solution

Answer

For both parts, the solution is \(x=1\).

Step 1: Use Logarithm Property

Recall the logarithmic property: \(\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)\). Apply this property to both sides: \(\log_2\left(\frac{x+5}{4x}\right) = \log_2(x)\).

Step 2: Equate the insides of the logarithms

Since the logarithmic function with the same base is one-to-one, equate the inside expressions: \(\frac{x+5}{4x} = x\).

Step 3: Solve the equation for x

Multiply both sides by \(4x\) to clear the fraction: \(x+5 = 4x^2\). Rearrange to form a quadratic equation: \(4x^2 - x - 5 = 0\).

Step 4: Factor the quadratic equation

Use factoring: \(4x^2 - 5x + 4x - 5 = 0\). Factoring gives us: \((4x+5)(x-1)=0\).

Step 5: Solve for possible x-values

Set each factor equal to zero: \(4x+5=0\) and \(x-1=0\). Solve for \(x\) to get \(x=-\frac{5}{4}\) and \(x=1\).

Step 6: Check for extraneous solutions

Since \(x\) must be positive for the original logarithm to be defined, discard \(x=-\frac{5}{4}\). The only solution is \(x=1\).

Step 7: Apply Logarithm Property to Part b

Similar to part a, use the property: \(\ln(m) - \ln(n) = \ln\left(\frac{m}{n}\right)\). Apply this: \(\ln\left(\frac{x+5}{4x}\right) = \ln(x)\).

Step 8: Equate and solve for x in Part b

Set the insides equal since the natural log function is one-to-one: \(\frac{x+5}{4x} = x\). Multiply both sides by \(4x\) and solve: \(x+5 = 4x^2\). Rearrange to yield: \(4x^2 - x - 5 = 0\).

Step 9: Solve the quadratic equation

Factor as \((4x+5)(x-1)=0\). Possible solutions are \(x=-\frac{5}{4}\) and \(x=1\).

Step 10: Check validity of solutions

Discard \(x=-\frac{5}{4}\) since \(x\) must be positive for natural log to be defined. The valid solution for part b is also \(x=1\).

Key Concepts

Logarithm Properties

Logarithmic properties are a set of rules that help simplify and solve equations involving logarithms. These rules make it easier to work with logarithms when you're faced with problems needing simplification, as seen in the exercise. A key property used in solving the exercise is:

• \( \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \). This formula shows how subtraction of two logarithms can be converted into a single log with division inside.
• To use this property, ensure that the logarithms have the same base. In the exercise, the property is used for both parts: base 2 for part a and the natural base (e) for part b.

By applying logarithmic properties effectively, we can transform complex log expressions into simpler forms, thus making the equations much easier to solve.

Natural Logarithms

Natural logarithms are logarithms with base \( e \), where \( e \) is approximately equal to 2.718. They are denoted by \( \ln \), instead of saying \( \log_e \). Natural logs play a crucial role in many mathematical applications, especially in calculus and differential equations. In part b of the exercise, natural logs are used:

• The property \( \ln(m) - \ln(n) = \ln\left(\frac{m}{n}\right) \) is employed. This mirrors the same property seen with other logarithmic bases.
• Just like with base 2 logs, the natural logarithm uses the one-to-one property: if \( \ln(a) = \ln(b) \), then \( a = b \).

Recognizing when to use natural log properties can simplify tasks dramatically and help solve equations efficiently. This mirrors the process shown for solving equations using regular logarithms.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations often involves finding the roots or solutions for \( x \) that make the equation true.

• In the exercise, after using logarithmic properties and equating expressions, a quadratic equation of the form \( 4x^2 - x - 5 = 0 \) is obtained.
• There are several methods to solve quadratic equations: factoring, using the quadratic formula, or completing the square. In this exercise, factoring is used.
• Once the quadratic is factored as \((4x+5)(x-1)=0\), the zero product property is used, which states if \((p \cdot q = 0)\), then either \(p = 0\) or \(q = 0\).

Quadratic equations are a fundamental concept in algebra, and understanding how to solve them is crucial in both simple and complex mathematics problems.