Problem 38 - Algebra for College Students

Question

Evaluate each logarithmic expression. $$ \log _{2}\left(\log _{4} 16\right) $$

Step-by-Step Solution

Answer

The expression evaluates to 1.

Step 1: Solve the Inner Logarithm

First, we need to evaluate \( \log_4 16 \). We can re-express 16 as a power of 4. We know that 16 can be expressed as \( 4^2 \). Thus, \( \log_4 16 = \log_4 (4^2) \). By the properties of logarithms, this simplifies to 2 since \( \log_b (b^m) = m \). Therefore, \( \log_4 16 = 2 \).

Step 2: Solve the Outer Logarithm

Now that we have \( \log_4 16 = 2 \), we substitute this into the outer expression: \( \log_2 (\log_4 16) = \log_2 2 \). We know that \( 2 = 2^1 \). Using the property of logarithms \( \log_b(b^m) = m \), we find that \( \log_2 2 = 1 \).

Step 3: Conclude the Evaluation

The expression \( \log_2(\log_4 16) \) evaluates to \( \log_2 2 \), which is 1. Therefore, the entire logarithmic expression simplifies to 1.

Key Concepts

Logarithmic Properties

Logarithmic properties are useful tools when it comes to simplifying complex logarithmic expressions. They help us rewrite these expressions in simpler forms, making calculations much easier. These properties include:

• The product property: \( \log_b (xy) = \log_b x + \log_b y \)
• The quotient property: \( \log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y \)
• The power property: \( \log_b (x^m) = m \log_b x \)
• Understanding that \( \log_b(b^m) = m \), which states that the logarithm of a base raised to a power is simply that power.

These properties allow us to break down complex expressions. In the exercise, we used the fact that \( \log_4(4^2) = 2 \) because of the power property, illustrating how these foundational rules are applied to simplify logarithms effectively.
Mastering these properties can unlock the ability to easily tackle even the most challenging logarithmic problems.

Change of Base Formula

The change of base formula is a significant tool in logarithms, particularly when dealing with bases that are not easily evaluated or when using a calculator. The formula is given by:\[\log_b a = \frac{\log_c a}{\log_c b}\]Here, \(b\) and \(c\) represent different bases, and \(a\) is the argument of the logarithm. This formula allows you to change the base of a logarithm to a more convenient one, typically either base 10 (common logarithm) or base \(e\) (natural logarithm), so that it can be found using a standard calculator.For example, if you had to solve \(\log_4 16\) without knowledge of obvious powers, you could use the change of base formula:\[\log_4 16 = \frac{\log_{10} 16}{\log_{10} 4}\]Using this approach allows you to solve for logarithms even without mental calculations, by utilizing technologies readily available on hand. It highlights the flexibility of logarithms, showcase how they can be adapted to different mathematical scenarios.

Exponents in Logarithms

Exponents and logarithms are intrinsically related. Understanding this relationship is crucial when solving logarithmic expressions and plays a central role, as showcased in the exercise.The concept is based on the rule \( \log_b (b^m) = m \), expressing that the logarithm of a base raised to an exponent returns the exponent itself. This implicitly simplifies expressions such as \( \log_4 16 \), where 16 can be rewritten as \( 4^2 \), allowing for direct computation using the aforementioned rule.This rule arises because logarithms are the inverse of exponentiation, much like subtraction is the inverse of addition. When you apply a logarithm to a number that is a base raised to an exponent, the base and the environment cancel each other out, leaving just the exponent.This relationship suggests that, when faced with an expression like \( \log_p (p^x) \), it immediately simplifies to \( x \). Such fundamental principles pave the way to tackling logarithmic problems and adding to one's mathematical toolkit efficiently.