Problem 30 - Algebra 2

Question

Solve each equation. Check your solutions. \(2 \log _{10} 6-\frac{1}{3} \log _{10} 27=\log _{10} x\)

Step-by-Step Solution

Answer

The solution is \( x = 12 \).

Step 1: Apply Logarithm Power Rule

First, apply the logarithm power rule: \( a \cdot \log_b(c) = \log_b(c^a) \). This transforms \( 2 \log_{10} 6 \) into \( \log_{10}(6^2) \) which is \( \log_{10} 36 \). Similarly, \( \frac{1}{3} \log_{10} 27 \) becomes \( \log_{10}(27^{1/3}) \). Since \( 27^{1/3} \) is the cube root of 27, it simplifies to 3. Now, the equation becomes \( \log_{10} 36 - \log_{10} 3 = \log_{10} x \).

Step 2: Apply Logarithm Subtraction Rule

Use the subtraction property of logarithms: \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \). Substituting into the equation, we have \( \log_{10}\left(\frac{36}{3}\right) = \log_{10} x \). Simplify \( \frac{36}{3} \) to 12, so the equation is \( \log_{10} 12 = \log_{10} x \).

Step 3: Equate the Arguments

Since the bases and the expressions on both sides of the equation are the same, equate the arguments: \( 12 = x \).

Step 4: Check the Solution

Substitute \( x = 12 \) back into the original equation: \( 2 \log_{10} 6 - \frac{1}{3} \log_{10} 27 = \log_{10} 12 \). Calculate: \( 2 \log_{10} 6 = \log_{10} 36 \) and \( \frac{1}{3} \log_{10} 27 = \log_{10} 3 \). Then \( \log_{10} 36 - \log_{10} 3 = \log_{10}\left(\frac{36}{3}\right) = \log_{10} 12 \). Thus, the solution is verified.

Key Concepts

Logarithm Power Rule

The logarithm power rule is an essential concept that simplifies expressions where logarithms are involved. This rule states that the logarithm of a number raised to a power can be rewritten as the product of that power and the logarithm of the base number. Mathematically, it is expressed as:

• \( a \cdot \log_b(c) = \log_b(c^a) \)

This property allows us to "move" the exponent in front of the logarithm, making complex operations more manageable.
In our exercise, we applied this rule to transform the term \( 2 \log_{10} 6 \) into \( \log_{10}(6^2) \), which simplifies to \( \log_{10} 36 \). Similarly, by applying the power rule, \( \frac{1}{3} \log_{10} 27 \) becomes \( \log_{10}(27^{1/3}) \). Understanding this rule is crucial, as it lays the groundwork for further simplification and solving more intricate logarithmic equations.

Logarithm Subtraction Rule

The logarithm subtraction rule is another important property of logarithms that helps simplify expressions where subtraction is involved. This rule states that the difference between two logarithms with the same base can be rewritten as the logarithm of a quotient. It is given by:

• \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \)

This property allows us to combine or simplify logarithmic expressions, often making it easier to solve equations.
In the provided exercise, we used the subtraction rule to combine the logarithms \( \log_{10} 36 - \log_{10} 3 \) into \( \log_{10}\left(\frac{36}{3}\right) \). Simplifying the quotient, we found \( \log_{10} 12 \), which is crucial for equating it to \( \log_{10} x \) and finding the value of \( x \). This demonstrates the subtraction rule's power in solving log equations by transforming a subtraction into a division.

Cube Root

The concept of a cube root is another mathematical operation that often appears in logarithmic transformations and problems. The cube root of a number is the value that, when multiplied by itself three times, gives the original number. It can be represented as:

• \( 27^{1/3} = 3 \)

Finding the cube root is essential in various contexts, especially when transformations involving fractional exponents occur.
In our exercise, after applying the logarithm power rule on \( \frac{1}{3} \log_{10} 27 \), we needed to compute the cube root of 27, which results in 3. This step was crucial in simplifying the expression and moving forward with the equation solving process. Understanding how to find cube roots is key to dealing with logarithmic equations involving fractional exponents, making them more approachable.