/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Exer. \(17-34\) : Solve the equa... [FREE SOLUTION] | 91Ó°ÊÓ

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Exer. \(17-34\) : Solve the equation. $$\log _{4}(3 x+2)=\log _{4} 5+\log _{4} 3$$

Short Answer

Expert verified
The solution is \( x = \frac{13}{3} \).

Step by step solution

01

Understand Logarithmic Properties

Notice the equation given involves logarithms with the same base. We can use the property of addition of logarithms, which states \( \log_b A + \log_b B = \log_b (AB) \). This will help us simplify the right side of the equation.
02

Apply Logarithmic Addition Property

Apply the property to the right side of the equation: \( \log_4 5 + \log_4 3 = \log_4 (5 \times 3) \). This simplifies to \( \log_4 15 \). Now the equation is \( \log_4(3x + 2) = \log_4 15 \).
03

Set the Arguments Equal

Since both sides of the equation are now single logarithms with the same base, set their arguments equal to each other: \( 3x + 2 = 15 \).
04

Solve for x

To solve \( 3x + 2 = 15 \), subtract 2 from both sides: \( 3x = 13 \). Then divide by 3: \( x = \frac{13}{3} \).
05

Verify the Solution

Plug \( x = \frac{13}{3} \) back into the original arguments to ensure they match. \( 3 \left( \frac{13}{3} \right) + 2 = 15 \), which satisfies the equation. Therefore, \( x = \frac{13}{3} \) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Properties
Logarithmic properties are essential tools for simplifying and solving equations involving logarithms. These properties help us manipulate the expressions to make solving equations more manageable. One of the most crucial properties is the Product Rule, which we used in the original exercise. This rule states that if you add two logarithms of the same base, you can express it as a single logarithm of their product:\[ \log_b A + \log_b B = \log_b (AB) \]This becomes particularly useful when you have an equation with logarithms on both sides. Another property worth mentioning is the Quotient Rule, which helps when dealing with logarithm division, expressed as:\[ \log_b \left( \frac{A}{B} \right) = \log_b A - \log_b B \]Finally, the Power Rule allows you to take an exponent out of a logarithm, giving:\[ \log_b (A^C) = C \cdot \log_b A \]Understanding and applying these properties allow you to break down/solve otherwise complex logarithmic equations. By recognizing when and how to use them, you make solving much simpler.
Solving Equations
When solving equations involving logarithms, like in the exercise, a systematic approach can clarify the solution process. Start by ensuring all terms are logarithms with a common base.If the equation has logarithms of the same base on both sides, like in our example, you can set the arguments equal to each other after using appropriate logarithmic properties. This method stems from the principle that if \( \log_b X = \log_b Y \), then \( X = Y \). An example from the exercise is setting:\[ \log_4(3x + 2) = \log_4 15 \]This directly leads to:\[ 3x + 2 = 15 \]Then, solve the resulting basic algebraic equation. This usually involves the following steps:
  • Isolate the variable term by adding or subtracting constants.
  • Divide or multiply to solve for the variable.
Always check your solution by substituting it back into the original equation to ensure it satisfies all conditions.
Logarithmic Functions
Logarithmic functions are useful tools in mathematics, often used to model exponential growth or decay. A logarithmic function is the inverse of an exponential function. It is typically expressed as:\[ y = \log_b x \]Here, \( y \) is the exponent to which the base \( b \) must be raised to produce \( x \). The base \( b \) is a constant, with base 10 and base \( e \) being the most common (these are called common logarithms and natural logarithms, respectively). Understanding the characteristics of logarithmic functions can help in comprehending growth patterns and rates in applied sciences.Logarithmic functions have several unique properties:
  • Their domain is restricted to positive real numbers.
  • The range is all real numbers.
  • They pass through the point (1, 0), since any number to the zero power is 1.
  • Their graph is a reflection across the line \( y=x \) of their corresponding exponential function's graph.
These properties make logarithmic functions distinct and widely applicable in various real-world scenarios.
Precalculus
Precalculus is an essential branch of mathematics that serves as a foundation for calculus studies. It involves the exploration of various mathematical concepts, including functions, equations, and their properties, as seen in our logarithmic equation exercise. Precalculus bridges the gap between algebra, geometry, and calculus, providing the tools needed to understand complex mathematical phenomena. A key aspect of precalculus is the focus on functions. Students learn about different types of functions, like polynomial, rational, and logarithmic functions, and their transformations. Solving equations, understanding asymptotes, and analyzing graphs help provide insights into function behavior. Precalculus also emphasizes the importance of trigonometry, sequences, and series. Mastery of these topics equips students with the skills needed to tackle calculus concepts, such as limits, derivatives, and integrals. Students often use precalculus knowledge to solve real-life problems, applying mathematical reasoning to diverse fields such as physics, engineering, and economics. This foundational subject enables learners to explore deeper mathematical worlds, sparking curiosity and developing critical thinking skills.

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Most popular questions from this chapter

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