Problem 57 - Intermediate Algebra

Question

For Problems \(51-59\), you are given \(\log _{2} 5=2.3219\) and \(\log _{2} 7=2.8074\). Evaluate each expression using Properties 11.5-11.7. $$ \log _{2} 175 $$

Step-by-Step Solution

Answer

\( \log_{2} 175 = 7.4512 \)

Step 1: Express the Expression using Logarithm Property

We begin by expressing \( 175 \) in terms of its prime factors. We know that \( 175 = 5 imes 5 imes 7 = 5^2 imes 7 \). Therefore, we can rewrite \( \log_{2} 175 \) using the logarithmic identities:\[ \log_{2} (5^2 \times 7) = \log_{2} (5^2) + \log_{2} (7) \]

Step 2: Apply Logarithm Power Property

Using the property of logarithms that states \( \log_b (x^n) = n \log_b (x) \), we break down \( \log_{2} (5^2) \):\[ \log_{2} (5^2) = 2 \log_{2} (5) \]

Step 3: Input Known Values

Now substitute the known values. We are given \( \log_{2} 5 = 2.3219 \) and \( \log_{2} 7 = 2.8074 \):- \( 2 \log_{2} 5 = 2 \times 2.3219 = 4.6438 \)- \( \log_{2} 7 = 2.8074 \) (as given)

Step 4: Combine the Results

Now we add the values we have found:\[ \log_{2} (5^2) + \log_{2} 7 = 4.6438 + 2.8074 \]

Step 5: Final Calculation

Perform the addition to get the final result:\[ 4.6438 + 2.8074 = 7.4512 \]

Step 6: Solution Conclusion

Thus, the evaluated expression \( \log_{2} 175 \) is equal to \( 7.4512 \).

Key Concepts

Logarithm Properties

Logarithms come with core properties that simplify complex calculations. These properties allow us to break down log expressions into simpler parts or combine them into a single log. Understanding these properties helps in evaluating expressions like \( \log_{2} 175 \).

• Product Property: This property is expressed as \( \log_b (xy) = \log_b x + \log_b y \). It allows us to split a log of a product into the sum of logs.


• Quotient Property: For division, \( \log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y \). It separates the log of a quotient into the difference of logs.


• Power Property: This property \( \log_b (x^n) = n \log_b x \) helps simplify log expressions involving exponents by extracting the exponent.

In the exercise, we applied the Product and Power properties. We transformed \( \log_{2} 175 \) to a sum \( \log_{2} (5^2 \times 7) = \log_{2}(5^2) + \log_{2} 7 \). Then by using the Power Property, \( \log_{2}(5^2) = 2 \log_{2} 5 \), simplifying our calculations significantly.

Change of Base Formula

The change of base formula is a crucial tool for working with logarithms. It lets you solve logarithmic expressions when you lack direct log values for specific bases. This formula goes like this: \[ \log_b x = \frac{\log_k x}{\log_k b} \] where \( k \) is any positive number not equal to 1, common choices being \( 10 \) or \( e \) for log and natural log, respectively.

Though the change of base formula wasn't directly applied in the provided solution, it's useful if the expression had been in a base other than 2. You could easily convert it using a different base that you have values for, which helps streamline calculations when working with less common bases.
The change of base formula reinforces that logarithms are flexible and adaptable for various calculations, enabling conversion into more convenient forms.

Logarithmic Identities

Logarithmic identities are equations that hold true for all exponential values. These identities assist in simplifying and manipulating expressions. Some well-known logarithmic identities include:

• Logarithm of 1: \( \log_b 1 = 0 \) for any base \( b \).


• Logarithm of base: \( \log_b b = 1 \) because any number raised to 1 is itself.


• Inverse Property: If \( b^{\log_b x} = x \), then \( \log_b (b^y) = y \).

These rules make complex logarithmic expressions more manageable and are pivotal when simplifying or solving equations involving exponents and logs. In the problem, these identities indirectly support how we broke down \( \log_{2} 175 \) into simpler parts using known values. By understanding these identities, solving logarithmic problems becomes more intuitive, allowing solutions to emerge more naturally.