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Chapter 3: Problem 30

Use the One-to-One Property to solve the equation for \(x\). $$\log _{2}(x-3)=\log _{2} 9$$

Short Answer

Expert verified
The solution to the equation is \(x = 12\)

Step by step solution

01

Apply the One-To-One Property of logarithms

According to the One-To-One Property of logarithms, if \(\log_b(m) = \log_b(n)\), then \(m=n\). Here the equation is \(\log_2(x-3) = \log_2(9)\). Hence after applying the One-To-One Property we get \(x-3=9\)
02

Solve for \(x\)

To solve for \(x\), add 3 to both sides of the equation to isolate x. The equation becomes \(x = 9 + 3\)
03

Addition

After performing the addition \(9 + 3\), \(x = 12\).

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Key Concepts

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

Understanding the One-to-One Property
The One-to-One Property is a fundamental concept of logarithms. This property states that if two logarithms with the same base are equal, then their arguments must be equal. This can be expressed as:
  • If \( \log_b(m) = \log_b(n) \), then \( m = n \).
In our exercise, we have \( \log_2(x-3) = \log_2(9) \). Since the base (2) is identical, we can conclude that the arguments must be equal. This simplifies our equation to \( x-3 = 9 \).
Using this property allows us to bypass complex calculations and solve the equation directly by equating the insides of the logarithms. This is a powerful tool when working with logarithmic equations.
Solving Equations Step-by-Step
Once we apply the One-to-One Property and simplify to \( x-3 = 9 \), our task is to solve for \( x \). Solving equations involves isolating the variable. Here鈥檚 how we do it:
  • We need to get \( x \) by itself on one side of the equation.
  • Add 3 to both sides to cancel out the \(-3\) from the left side.
After adding 3, the equation becomes \( x = 9 + 3 \).
Performing the addition gives us \( x = 12 \). This method isolates \( x \) by performing inverse operations, which are key to solving linear equations efficiently.
Exploring Properties of Logarithms
Logarithms have several useful properties which can simplify problem-solving:
  • The Product Property: \( \log_b(mn) = \log_b(m) + \log_b(n) \).
  • The Quotient Property: \( \log_b\left( \frac{m}{n} \right) = \log_b(m) - \log_b(n) \).
  • The Power Property: \( \log_b(m^n) = n \cdot \log_b(m) \).
These properties are essential in solving and simplifying logarithmic expressions and equations. In our exercise, we primarily used the One-to-One Property. However, understanding these additional properties can help tackle more complex equations down the line, providing flexibility in mathematical problem-solving.

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Most popular questions from this chapter

Condense the expression to the logarithm of a single quantity. $$2 \ln 8+5 \ln (z-4)$$

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562, \log _{b} 3 \approx 0.5646,\) and \(\log _{b} 5 \approx 0.8271.\) $$\log _{b}(2 b)^{-2}$$

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln \left(\frac{1}{x}\right)-x=0$$

Determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$f(0)=0$$

The values \(y\) (in billions of dollars) of U.S. currency in circulation in the years 2000 through 2010 can be modeled by \(y=-611+507 \ln t, 10 \leq t \leq 20\) where \(t\) represents the year, with \(t=10\) corresponding to 2000. During which year did the value of U.S. currency in circulation exceed \(\$ 690\) billion?
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