Problem 51 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each logarithmic equation. 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. $$\log _{3}(x-4)=-3$$

Step-by-Step Solution

Answer

The exact solution is \( x = 4 + \frac{1}{27} \). The decimal approximation to the solution, correct to two decimal places, is \( x \approx 4.04 \).

Step 1: Apply the definition of the logarithm

First, rearrange the logarithmic equation in exponent form. According to the definition of a logarithm, if \( a = \log_{b}(c) \), then \( b^a = c \). Therefore, our equation \( \log_{3}(x-4)=-3 \) can be rewritten as \( 3^{-3} = x - 4 \).

Step 2: Solving for x

Simplify and solve the equation for \( x \). To simplify \( 3^{-3} \), remember that the negative exponent means taking the reciprocal, thus \( 3^{-3} = \frac{1}{3^3} = \frac{1}{27} \). So the equation becomes: \( \frac{1}{27} = x - 4 \). Solve this equation by adding \( 4 \) to both sides to isolate \( x \). Therefore, \( x = 4 + \frac{1}{27} \).

Step 3: Convert the exact solution to decimal form

The exact solution is \( x = 4 + \frac{1}{27} \). To get the decimal approximation, calculate \( 4 + \frac{1}{27} \) using a calculator and round your answer to two decimal places. This gives approximately \( x \approx 4.04 \).

Step 4: Check the solution

Validate the solution by substituting it back into the original logarithmic equation. Plug \( 4.04 \) into the original equation \( \log_{3}(x-4)=-3 \) and check if it holds. A quick inspection shows the solution worked out well, so no value requires to be rejected.

Key Concepts

Logarithm Properties

Understanding logarithm properties is crucial when tackling equations like \(\log _{3}(x-4)=-3\). Logarithms, the inverse operations of exponentiation, have unique properties that allow us to manipulate and solve logarithmic equations easily.

One key property is the product rule, which states that \(\log_b(mn) = \log_b(m) + \log_b(n)\), letting us separate the logarithm of a product into sum of logarithms. Another important property is the quotient rule, \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\), used for dividing terms within a logarithm.

The power rule, \(\log_b(m^n) = n\cdot \log_b(m)\), lets us move an exponent outside the log function. Also, we have the change of base formula, allowing us to transform a log to a different base: \(\log_b(m) = \frac{\log_k(m)}{\log_k(b)}\) where \(k\) is a new base. Finally, the base-switching property, states that \(b^{\log_b(m)} = m\) and \(\log_b(b^m) = m\), express the inverse nature of logarithms and exponentiation.

These properties help us in rewriting and solving logarithmic equations and understanding the relationship between logarithms and exponents.

Exponent Form Conversion

When faced with a logarithmic equation such as \(\log _{3}(x-4)=-3\), converting to exponent form makes it much easier to solve. This process uses the basic definition of logarithms: if \(a = \log_{b}(c)\), then \(b^a = c\).

For our given equation, we transform \(\log _{3}(x-4)\) into \(3^{-3}\) by setting \(3\) as the base and \(x-4\) as the result. This conversion takes advantage of the inverse relationship between logarithms and exponents, turning a potentially complex logarithmic equation into a more familiar exponential form.

Solving After Conversion

Once in exponential form, solving for \(x\) becomes straightforward. In our problem, this resulted in the equation \(\frac{1}{27} = x - 4\), which we then solved for \(x\). This approach simplifies the calculation by using basic algebraic manipulations.

Domain of Logarithmic Functions

The domain of logarithmic functions is an essential consideration that can't be ignored while solving logarithmic equations. A log function like \(\log_b(x)\) is defined only for positive values of \(x\), because you can't take the log of a negative number or zero in real numbers.

In the exercise \(\log _{3}(x-4)=-3\), it's important to ensure that \(x-4 > 0\), otherwise the logarithmic expression would be undefined. After solving the equation step by step, we found \(x = 4 + \frac{1}{27}\), which is clearly greater than 4. Hence, it falls within the permissible domain of the function, and we can be confident that our solution is valid in the context of the original logarithmic expression.

Always check that the solved value does not cause the argument of the logarithm (the value inside the logarithmic function) to be zero or negative, as this would mean the solution is outside the function's domain and therefore not a valid solution to the equation.