Problem 54 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{5}(x-7)=2 $$

Step-by-Step Solution

Answer

The solution to the equation \(\log _{5}(x-7)=2\) is \(x = 32\).

Step 1: Understand Logarithm Properties

We must understand that \(\log _{a} b = c\) is equivalent to \(a^c = b(\)base \(a\) raised to the power of \(c\) equals \(b\)). So we can convert \(\log _{5}(x-7)=2\) to exponential form.

Step 2: Convert Logarithm to Exponential Form

By using the property, we change \(\log _{5}(x-7) =2\) into \(5^2= x-7\). It becomes \(25 = x-7\).

Step 3: Solve for \(x\)

We solve for \(x\) by adding \(7\) to both sides of the equation to isolate \(x\), so \(x = 25+7 = 32\).

Step 4: Check the Domain

We should verify that \(x = 32\) is in the domain of the original logarithmic expressions. In this case, for the logarithm to be real, \(x - 7 > 0\). Substituting \(x = 32\), we find that \(32 - 7 > 0\) is true. Thus, \(x = 32\) is in the domain and is therefore the solution.

Key Concepts

Exponential Form

The concept of exponential form is closely tied to logarithms. When dealing with logarithmic equations, understanding how to convert them to exponential form is crucial for solving them. In the given exercise, the equation is \( \log_{5}(x-7) = 2 \). This reads as "log base 5 of \( x-7 \) equals 2."

The exponential form takes advantage of the basic logarithm property: \( \log_{a} b = c \) translates to \( a^c = b \). Therefore, we rewrite \( \log_{5}(x-7) = 2 \) in exponential form as \( 5^2 = x - 7 \).

• "Log base 5" refers to the number 5 being the base in the exponential equation.
• "2" is the exponent to which the base is raised.
• The expression on the right, \( x - 7 \), is what the base and exponent are equated to.

By converting logarithmic equations to exponential form, we can often simplify the solving process, allowing us to solve for the variable more straightforwardly.

Properties of Logarithms

Understanding the properties of logarithms makes solving logarithmic equations more approachable. These properties are rules that simplify logarithmic expressions and manipulations. Let's highlight some essential concepts:

• Logarithm of a Product: \( \log_b(MN) = \log_b M + \log_b N \).
• Logarithm of a Quotient: \( \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \).
• Logarithm of a Power: \( \log_b(M^n) = n \cdot \log_b M \).
• Changing the Base: Allows changing the base of a logarithm, useful for computation, given by \( \log_b m = \frac{\log_k m}{\log_k b} \) where \( k \) is the new base.

These properties were not directly used in the exercise, but they are valuable tools for simplifying logarithmic equations in general, helping to align terms or break down complex expressions.

Domain of Logarithmic Functions

When solving logarithmic equations, the domain of the original logarithmic function is crucial to ensure solutions are valid. Logarithmic functions, like \( \log_b x \), have specific domain requirements:

• The expression inside the logarithm (the argument) must be greater than zero; it cannot be zero or negative.
• For the equation \( \log_5(x-7) \), this means \( x - 7 > 0 \), or \( x > 7 \).
• This restriction comes from the definition of logarithms, as you can't take a logarithm of a non-positive number.

In our exercise, we solved \( 5^2 = x - 7 \) finding \( x = 32 \). We then checked to ensure \( x = 32 \) met the domain requirement \( x - 7 > 0 \). Substituting 32 into the inequality \( 32 - 7 > 0 \) confirms the solution is valid and within the domain, thus solving our equation correctly.