Problem 82 - Precalculus with Calculus Previews

Question

Discuss: Are the given two equations equivalent, that is, do they have the same solution set? (a) \(\log _{5}(x-2)^{2}=2 ; 2 \log _{5}(x-2)=2\) (b) \(\log _{5}(x-2)^{3}=3 ; 3 \log _{5}(x-2)=3\)

Step-by-Step Solution

Answer

Both pairs of equations are equivalent; they both have the solution \( x = 7 \).

Step 1: Understanding Equivalence of Equations

Two equations are said to be equivalent if they have the same solution set. To determine this, we must solve each equation for their solutions and compare them.

Step 2: Solving Equation (a) Logarithmically

Given: \( \log _{5}(x-2)^{2}=2 \). By the properties of logarithms, this can be rewritten as \( (x-2)^{2} = 5^2 \). Solving this, we get \( x-2 = 5 \) or \( x-2 = -5 \), which gives the solutions \( x = 7 \) or \( x = -3 \). The solution \( x = -3 \) is invalid as it results in taking the log of a negative number. Thus, the valid solution is \( x = 7 \).

Step 3: Solving Equation (a) Using Logarithmic Identity

For the equation \( 2 \log _{5}(x-2) = 2 \), we divide each side by 2 to get \( \log _{5}(x-2) = 1 \). This implies \( x-2 = 5^1 \), leading to \( x = 7 \).

Step 4: Conclusion for Part (a)

Both equations in part (a) have the single solution \( x = 7 \), hence they are equivalent.

Step 5: Solving Equation (b) Logarithmically

Given: \( \log _{5}(x-2)^{3}=3 \). Rewriting, we have \( (x-2)^{3} = 5^3 \). Solving this results in \( x-2 = 5 \), leading to the solution \( x = 7 \).

Step 6: Solving Equation (b) Using Logarithmic Identity

For \( 3 \log _{5}(x-2) = 3 \), divide both sides by 3 to get \( \log _{5}(x-2) = 1 \). This gives \( x-2 = 5^1 \), resulting in \( x = 7 \).

Step 7: Conclusion for Part (b)

Both equations in part (b) have the single solution \( x = 7 \), therefore they are equivalent.

Step 8: Final Conclusion

For parts (a) and (b), the equations are equivalent as they both have the same solution set, \( x = 7 \).

Key Concepts

Logarithmic Equations

Logarithmic equations are equations that involve logarithms with variables inside them. These types of equations can be solved by applying the properties of logarithms, such as turning an equation into its exponential form, which often simplifies the process. In the exercise provided, we start with logarithmic equations like \( \log_{5}(x-2)^{2} = 2 \). Here's what to remember when dealing with these equations:

• The base of the logarithm plays a significant role. In this case, it's 5.
• When you have a logarithm set equal to a number, you can often rewrite it in an exponential form. For example, \( \log_{5}(x-2)^{2} = 2 \) becomes \( (x-2)^{2} = 5^{2} \).
• Solving the resulting exponential equation gives you potential values for \(x\).

Remember, logarithms are useful for solving equations where you need to "undo" an exponent or find the exponent in equations. Always check your solutions, because some might not work, especially if they lead to taking a logarithm of a negative number.

Solution Sets

The solution set of an equation is the set that includes all values that satisfy the equation, meaning that when substituted back into the equation, they maintain its validity. In this exercise, we determined the solution sets for the logarithmic equations by solving them step by step.Let's break it down:

• For a given logarithmic equation, solve it and list any possible solutions. Often, you'll get more than one solution before checking their validity.
• Check the potential solutions by substituting them back into the original equation. If a solution results in an invalid operation, like the logarithm of a negative number, it should not be part of the final solution set.
• The final list of valid solutions forms the solution set for that equation.

In the problem addressed, only \(x = 7\) is valid for all given equations, meaning the solution set for each is \(\{7\}\). Always ensure your solution set contains only feasible solutions for the type of equation you're working with.

Properties of Logarithms

Understanding the properties of logarithms is crucial for solving logarithmic equations efficiently. These properties allow us to manipulate logarithmic expressions to make them easier to analyze and solve.Here are some key properties:

• Product Property: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• Power Property: \( \log_b(M^k) = k \cdot \log_b(M) \)

In the original exercise, the power property is particularly relevant. For instance, \( \log_{5}(x-2)^{2} \) can be rewritten using the power property as \( 2 \log_{5}(x-2) \). This simplification helps to relate and manipulate the equations further. Knowing and using these properties not only simplifies the problem but also maintains the integrity of equations when solving them. They're the foundational tools needed when working with any logarithmic equation.