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Chapter 12: Problem 41

Solve each equation. Give exact solutions. \(\log (6 x+1)=\log 3\)

Short Answer

Expert verified
x = \frac{1}{3}

Step by step solution

01

- Understand the Problem

The problem requires solving the equation \( \log (6x + 1) = \log 3 \) for exact values of x.
02

- Use the Property of Logarithms

If \( \log A = \log B \), then \( A = B \). Apply this property to the given equation: \( 6x + 1 = 3 \).
03

- Solve for x

Isolate x by first subtracting 1 from both sides: \( 6x + 1 - 1 = 3 - 1 \, which simplifies to \, 6x = 2 \). Then, divide both sides by 6 to get \, x = \frac{2}{6} \. Simplify \, x = \frac{1}{3} \.

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Key Concepts

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

headline of the respective core concept
When solving logarithmic equations like \(\log \(6x + 1\) = \log 3\), the key is to understand and apply the properties of logarithms correctly. Logarithmic equations may initially seem daunting, but they become more manageable by breaking them down into simpler steps.
First, we need to understand that if \(\log A = \log B\), then their arguments must be equal, which means simply \(A = B\). In this case, \(A\) is \(6x + 1\) and \(B\) is 3.
By eliminating the logarithms using this property, we transform the problem into a straightforward algebraic equation: \(6x + 1 = 3\). This approach makes our task much simpler.
headline of the respective core concept
Understanding the fundamental properties of logarithms is crucial. The primary property we use in our problem is: if \(\log A = \log B\), then \(A = B\). This property allows us to remove the logarithms and work only with the arguments.
In addition to this key property, it's useful to remember:
  • Product Rule: \(\log_b(AB) = \log_b A + \log_b B\)
  • Quotient Rule: \(\log_b\left(\frac{A}{B}\right) = \log_b A - \log_b B\)
  • Power Rule: \(\log_b\left(A^n\right) = n \log_b A\)
Although these properties aren't applied in the given exercise, they are fundamentally important for solving a variety of logarithmic equations you might encounter in the future. Keeping these rules in mind will provide a toolkit to navigate through diverse logarithmic problems.
headline of the respective core concept
In our example, we derived the exact solution by isolating the variable x after eliminating the logarithms. Setting up \(6x + 1 = 3\) and solving for x through simple algebra is the next step.
First, subtract 1 from both sides to isolate the term containing x:
  • \(6x + 1 - 1 = 3 - 1\)
  • which simplifies to \(6x = 2\)
Next, solve for x by dividing both sides by 6:
  • \(x = \frac{2}{6}\)
  • which simplifies to \(x = \frac{1}{3}\)
Thus, the exact solution to the logarithmic equation \(\log \(6x + 1\) = \log 3\) is \(x = \frac{1}{3}\). This step-by-step process emphasizes the importance of breaking down algebraic manipulations into manageable stages. Mastery of these techniques allows for tackling even more challenging equations with confidence.

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Most popular questions from this chapter

To four decimal places, the values of \(\log _{10} 2\) and \(\log _{10} 9\) are $$\log _{10} 2=0.3010 \text { and } \log _{10} 9=0.9542$$ Use these values and the properties of logarithms to evaluate each expression. DO NOT USE A CALCULATOR. $$\log _{10} 36$$

Solve each equation. Give exact solutions. \(\log _{4}(2 x+8)=2\)

Solve each equation. Use natural logarithms. Approximate solutions to three decimal places when appropriate. $$ e^{0.012 x}=23 $$

Use the properties of logarithms to express each logarithm as a sum or difference of logarithms, or as a single logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{3} \frac{\sqrt[3]{4}}{x^{2} y}$$

Suppose that you are an agent for a detective agency. Today's encoding function is \(f(x)=4 x-5 .\) Find the rule for \(f^{-1}\) algebraically.
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