Problem 31 - College Algebra

Question

Solve each logarithmic equation in Exercises \(27-44 .\) Be sure to reject any value of \(x\) that produces the logarithm of a negative number or the logarithm of \(0 .\) $$\log _{3}(x-4)=-3$$

Step-by-Step Solution

Answer

The solution to \(\log _{3}(x-4)=-3\) is \(x = \frac{28}{27}\)

Step 1: Convert the logarithm to exponential form

To convert the given logarithmic equation \(\log _{3}(x-4)=-3\) into exponential form you apply the conversion rule, which says: the \(b\) raised to the power of \(y\) equals \(x\) (where \(b\) is the base, \(y\) is the exponent, and \(x\) is the argument) . This gives \(3^{-3} =x-4\)

Step 2: Solve for \(x\)

Next, solve the equation for \(x\). This is done by adding \(4\) to both sides of the equation: \(x = 3^{-3}+4 \)

Step 3: Calculate \(x\)

Calculate the value of \(x\) by substituting \(3^{-3}\) into the equation, and remember the rule of negative exponent. It gives \(x = \frac{1}{3^3}+4 = \frac{1}{27}+4 = \frac{28}{27}\)

Step 4: Check if value is valid

The solution to the logarithmic equation is \(x = \frac{28}{27}\). However, we must also verify that this value does not make the argument of the given logarithm non-positive. Substitute \(x\) back into the equation and it gives \(\log_{3}(\frac{28}{27} - 4)\). This argument is positive, thus the solution is valid

Key Concepts

Logarithmic to Exponential Form

Understanding how to transform a logarithmic equation into its exponential form is crucial for solving logarithmic equations. The general rule to convert a logarithm ewline \(log_b(a) = c\) to exponential form is by expressing it as \(b^c = a\). In this process, 'b' represents the base of the logarithm, 'c' is the value to which the base is raised, and 'a' is the result of this exponential expression.

For example, let's consider the equation ewline \(log_{3}(x-4)=-3\). By applying the conversion rule, the equation can be rewritten in exponential form as ewline \(3^{-3} = x - 4\). This step is the foundation for finding the value of 'x' that satisfies the given logarithmic equation. By using the exponential form, we can apply familiar algebraic techniques to solve for 'x'.

Logarithm Properties

Logarithms have unique properties that aid in simplifying complex equations. Some of these properties are based on the fundamental nature of logarithms and how they relate to exponents. Here are a few key properties:

• Product Rule: \(log_b(mn) = log_b(m) + log_b(n)\), which means the logarithm of a product is the sum of the logarithms.
• Quotient Rule: \(log_b\left(\frac{m}{n}\right) = log_b(m) - log_b(n)\), which tells us the logarithm of a quotient is the difference between the logarithms.
• Power Rule: \(log_b(m^k) = k \cdot log_b(m)\), indicating that the logarithm of a power is the exponent times the logarithm.
• Change of Base Rule: This allows the conversion of logarithms from one base to another, which can be formulated as \(log_b(a) = \frac{log_k(a)}{log_k(b)}\) for any positive base 'k'.

Knowing and applying these properties appropriately can help simplify and solve logarithmic equations more effectively. They provide a powerful toolkit for manipulating equations to a form that is easier to deal with, especially when it comes to complex problems.

Checking Logarithmic Solutions

It's crucial not to overlook the final step after solving a logarithmic equation, which is verifying that your solution is correct. When we find a value for 'x' in a logarithmic equation, that value must result in a positive number when substituted back into the original logarithm. This is because the logarithm of a negative number or zero is undefined in real numbers.

In our example, after solving for 'x', we obtained \( x = \frac{28}{27}\). To confirm its validity should be part of your routine: You substitute \( x\) back into the original logarithmic expression, ewline \(log _{3}(x-4)\). For our solution, substituting gives us: ewline \(log_{3}(\frac{28}{27} - 4)\), which simplifies to ewline \(log_{3}(\frac{1}{27})\).Since \(\frac{1}{27}\) is positive, we conclude that \( x = \frac{28}{27}\) is indeed a valid solution. Remember to always test your answer to avoid including solutions that don't actually satisfy the original equation.