Problem 66 - Algebra for College Students

Question

Given that \(\log _{8} 5=0.7740\) and \(\log _{8} 11=1.1531\), evaluate each expression using Properties \(10.5-10.7\) \(\log _{8} 320\)

Step-by-Step Solution

Answer

\(\log_{8} 320 = 2.7740\).

Step 1: Understand the Expression

The expression we need to evaluate is \(\log_{8} 320\). We need to express 320 in terms of known values or factor it appropriately for ease of calculation using known logarithmic properties.

Step 2: Factor 320

Factor 320 into primes or numbers related to 5 and 11 since we have the logs for both: \(320 = 5 \times 64 = 5 \times 8^2\).

Step 3: Apply Logarithm Product Rule

Using the property \(\log_{b}(mn) = \log_{b}m + \log_{b}n\), rewrite: \(\log_{8} (5 \times 8^2) = \log_{8} 5 + \log_{8} 8^2\).

Step 4: Calculate Each Term

First term is: \(\log_{8} 5 = 0.7740\). Second term is: \(\log_{8} 8^2 = 2 \cdot \log_{8} 8 = 2 \cdot 1 = 2\) because \(\log_{8} 8 = 1\).

Step 5: Combine the Values

Now sum the two logarithmic values to find \(\log_{8} 320\). Hence, \(\log_{8} 320 = 0.7740 + 2 = 2.7740\).

Key Concepts

Logarithmic Properties

Logarithms are a mathematical tool that help us solve exponential equations by transforming multiplication into addition. This is possible due to various logarithmic properties. One such property is the product rule, which states:

• \(\log_{b}(mn) = \log_{b}m + \log_{b}n\)

This property is key to breaking down complex logarithmic expressions into simpler, more manageable parts that we already know. The base of the logarithm, represented by \(b\), remains constant while the expression itself transforms.
For instance, in our exercise, we used the product rule to evaluate \(\log_{8} (5 \times 8^2)\), breaking it into parts for which the logarithms were already given or could be easily calculated.
These properties also make logarithms a powerful method for calculating exponential growth, such as population dynamics or financial interest where direct calculation isn't straightforward.

Factorization

Factorization involves breaking down numbers into smaller components, often primes, that multiply to form the original number. This concept is crucial in rearranging mathematical expressions into forms that are easier to work with.
In our given exercise, factoring 320 as \(320 = 5 \times 8^2\) helped us link the problem back to known logarithmic values. By expressing 320 through its factors of converted logarithm bases (like 8 and its squared form), we simplify the task of evaluating the logarithm of 320 using known components \(\log_{8} 5\) and known identities like \(\log_{8} 8 = 1\).
This step is vital in simplifying logarithmic expressions, ensuring that they can be approached with elementary log values or previously calculated logarithms, thereby easing the calculation process.

Product Rule

The product rule is a fundamental property of logarithms, which allows us to break down an expression with a product inside the logarithm into the sum of two separate logarithms. The rule is expressed as follows:

• \(\log_{b}(mn) = \log_{b}m + \log_{b}n\)

This rule greatly simplifies calculations by allowing the evaluation of complex expressions using simpler known parts. In the exercise, we took advantage of the product rule to transform the expression \(\log_{8}(5 \times 8^2)\) into \(\log_{8}5 + \log_{8}8^2\), resulting in easier computations.
The product rule is especially useful in handling multiplicative situations in real applications, such as combining different growth rates or consolidating product yields in scientific data, transforming otherwise complex equations into tractable forms.

Change of Base

The change of base formula is a useful tool in logarithms when we need to compute a logarithm with a base that is not readily available. It allows us to convert a logarithm from one base to another known base. The change of base formula is given by:

• \(\log_{b} a = \frac{\log_{k} a}{\log_{k} b}\)

This formula is particularly helpful when using calculators, as most are configured to calculate logarithms only in base 10 or base \(e\). By transforming our calculations to a usable base, we can leverage standard technology to solve problems that would otherwise be difficult to compute by hand.
While the exercise didn't require the change of base formula directly, understanding this concept offers additional flexibility when approaches involving known logarithmic values don't suffice, securing confidence in tackling a wider range of problems.