Problem 91 - Intermediate Algebra

Question

Assume that \(\log 4 \approx 0.6021, \log 7 \approx 0.8451,\) and \(\log 9 \approx 0.9542 .\) Use these values to evaluate each logarithm. See Example 8 . $$ \log _{b} \frac{63}{4} $$

Step-by-Step Solution

Answer

The value of \( \log_{b} \left( \frac{63}{4} \right) \approx 1.1972 \).

Step 1: Express the Logarithmic Expression

Begin by expressing the logarithm \( \log_{b} \frac{63}{4} \) using the properties of logarithms. This can be rewritten as the difference of two logarithms using the quotient rule: \[ \log_{b} \left( \frac{63}{4} \right) = \log_{b} (63) - \log_{b} (4) \]

Step 2: Decompose Logarithm with Prime Factors

Next, express the number 63 in terms of its prime factors: \(63 = 7 \times 9 \). Thus, \( \log_{b} (63) = \log_{b} (7 \times 9) \).

Step 3: Apply the Product Rule

Use the product rule of logarithms to decompose \( \log_{b}(7 \times 9) \) into a sum: \[ \log_{b}(7 \times 9) = \log_{b}(7) + \log_{b}(9) \]

Step 4: Substitute the Known Values

Substitute the given logarithm values into the expression: \( \log_{b}(7) \approx 0.8451 \), \( \log_{b}(9) \approx 0.9542 \), \( \log_{b}(4) \approx 0.6021 \). Thus, we have: \[ \log_{b}(63) = \log_{b}(7) + \log_{b}(9) \approx 0.8451 + 0.9542 \approx 1.7993 \]

Step 5: Complete the Calculation

Substitute \( \log_{b}(63) \) back into the expression from Step 1: \[ \log_{b} \left( \frac{63}{4} \right) = \log_{b} (63) - \log_{b} (4) \approx 1.7993 - 0.6021 \] Calculate the result: \[ \log_{b} \left( \frac{63}{4} \right) \approx 1.1972 \]

Key Concepts

Quotient Rule

When dealing with logarithms, the quotient rule is a very handy tool. It helps to simplify expressions where you have a division inside the logarithm. The rule states that the logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator.
So, if you come across an expression such as \( \log_{b} \left( \frac{A}{B} \right) \), you can rewrite it as \( \log_{b}(A) - \log_{b}(B) \). This makes solving logarithmic problems much easier, especially when you know the logarithm values of \( A \) and \( B \).
In our exercise, we applied this rule to break down \( \log_{b} \left( \frac{63}{4} \right) \) into simpler parts: \( \log_{b}(63) - \log_{b}(4) \). By doing so, we can use known values to arrive at the solution.

Product Rule

Another important property of logarithms is the product rule. This is particularly useful when you need to handle a product inside a logarithm. The product rule tells us that the logarithm of a product is the sum of the logarithms of the factors.
Mathematically, this is expressed as \( \log_{b}(A \times B) = \log_{b}(A) + \log_{b}(B) \). This property allows us to break down complicated expressions into simpler parts, making calculations more manageable.
In the exercise, we used the product rule to decompose \( \log_{b}(63) \) into \( \log_{b}(7 \times 9) \) which then becomes \( \log_{b}(7) + \log_{b}(9) \). Knowing the values of \( \log_{b}(7) \) and \( \log_{b}(9) \) allowed us to easily compute \( \log_{b}(63) \). This step is crucial for handling more complex logarithmic expressions.

Prime Factorization

Prime factorization is a powerful technique used to break down a number into its basic building blocks - prime numbers. Every integer greater than 1 can be represented uniquely as a product of prime numbers, which is essential in simplifying expressions involving logarithms.
Consider the number 63 from the exercise. We express it using its prime factors: \( 63 = 7 \times 9 \), where 7 is a prime number and 9 can be further broken down as \( 3 \times 3 \) or \( 3^2 \). In our case, using \( 7 \times 9 \) was sufficient, as the known values for \( \log_{b}(7) \) and \( \log_{b}(9) \) were given.
This step allowed us to simplify the logarithm \( \log_{b}(63) \) using the product rule. Using prime factors can make complex logarithmic evaluations much easier by reducing larger numbers to their simplest multiplicative form.