Problem 40 - Algebra for College Students

Question

Evaluate each logarithmic expression. $$ \log _{2}\left(\log _{5} 5\right) $$

Step-by-Step Solution

Answer

The value of the logarithmic expression is 0.

Step 1: Evaluate Inner Logarithm

Start by evaluating the inner logarithmic expression, \(\log_{5}5\). Using the knowledge that \(\log_{b}b = 1\) for any positive base \(b\), it follows that \(\log_{5} 5 = 1\).

Step 2: Substitute the Result into the Outer Logarithm

With the result from Step 1, substitute \(1\) into the outer logarithmic expression: \(\log_{2}(1)\).

Step 3: Evaluate the Outer Logarithm

Evaluate \(\log_{2}(1)\). Recall that for any positive base \(b\), \(\log_{b}1 = 0\). Therefore, \(\log_{2}(1) = 0\).

Key Concepts

Logarithmic Expressions

Logarithmic expressions involve expressions that contain a logarithm, which is essentially the inverse operation of exponentiation. This means they help us answer the question: "To what exponent must a base be raised to yield a certain value?" In the expression \(\log _{2}\left(\log _{5} 5\right)\), two logarithms are used, one inside the other. This is known as nesting. The inner logarithmic expression, \(\log _{5} 5\), asks us to find out to what power we must raise 5 to get 5. This is a relatively simple calculation due to the property that \(\log_{b}b = 1\). So, \(\log_{5}5 = 1\). By breaking down these expressions step by step, we can solve more complex problems effectively and gain a deeper understanding of logarithmic functions.
Logarithmic expressions are powerful tools in mathematics, engineering, and science, often used when working with exponential growth and decay. They help us express and solve equations that would otherwise be difficult to handle. Remembering basic logarithmic identities, such as \(\log_{b}b = 1\) and \(\log_{b}1 = 0\), is crucial when manipulating these expressions.

Properties of Logarithms

Understanding the properties of logarithms is essential when working with logarithmic expressions. These properties help simplify expressions and solve equations more efficiently. Some key properties include:

• Product Property: \(\log_b(MN) = \log_b M + \log_b N\)
• Quotient Property: \(\log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N\)
• Power Property: \(\log_b(M^k) = k \cdot \log_b M\)
• Change of Base Formula: \(\log_b M = \frac{\log_k M}{\log_k b}\) for any positive base \(k\)

These properties are highly useful for rewriting complicated logarithmic expressions into more manageable forms. For example, when solving \(\log_{2}(1)\), we apply the fundamental property that \(\log_{b}1 = 0\). This reminds us that any number's log with respect to 1, regardless of the base \(b\), is zero.
By mastering these properties, you can tackle a broad range of logarithmic problems. Always remember that these rules can be combined and adapted to fit the varying structures of logarithmic equations you might encounter.

Logarithm Evaluation

The process of evaluating a logarithm involves determining the exponent that a given base must be raised to obtain a specific number. When evaluating logarithmic expressions, it is important to approach them systematically. In the expression from the exercise, \(\log_{2}(\log_{5}5)\), we worked with a nested logarithmic expression. First, calculate the inner expression, knowing that \(\log_{5}5 = 1\) because 5 to the power of 1 equals 5. This simplifies the expression to \(\log_{2}(1)\).
Next, move to the outer expression. It then requires you to determine what power of 2 will yield 1. Since any base raised to the power of 0 is 1, \(\log_{2}(1) = 0\).
Evaluating logarithms step by step, especially in compound expressions, becomes manageable when you understand both what the logarithms ask and the properties they adhere to. Practice with these processes will enhance your ability to solve complex problems systematically.