In Exercises 105–108, evaluate each expression without using a calculator. log₅ (log₂ 32)

Verified step by step guidance

1. Identify the expression to evaluate: \( \log_5 \left( \log_2 32 \right) \). This means we first need to evaluate the inner logarithm \( \log_2 32 \) before applying the outer logarithm.
2. Evaluate the inner logarithm \( \log_2 32 \). Recall that \( \log_b a = c \) means \( b^c = a \). So, find the exponent \( c \) such that \( 2^c = 32 \).
3. Since \( 32 = 2^5 \), it follows that \( \log_2 32 = 5 \). Now substitute this value back into the original expression to get \( \log_5 5 \).
4. Evaluate the outer logarithm \( \log_5 5 \). Using the definition of logarithms, \( \log_b b = 1 \) for any base \( b > 0 \) and \( b \neq 1 \).
5. Therefore, \( \log_5 5 = 1 \). This completes the evaluation of the original expression.

Key Concepts

Understanding Logarithms

A logarithm answers the question: to what exponent must the base be raised to produce a given number? For example, log_b(a) = c means b^c = a. This concept is fundamental for evaluating and simplifying logarithmic expressions.

Evaluating Inner Logarithmic Expressions

When dealing with nested logarithms like log5(log2 32), first evaluate the inner logarithm. Recognize powers of numbers, such as 32 = 2^5, to simplify log2 32 to 5, which then becomes the argument for the outer logarithm.

Properties of Logarithms and Simplification

Using properties like log_b(b) = 1 helps simplify expressions. After evaluating the inner log, apply these properties to the outer log. This stepwise simplification avoids calculators and relies on understanding logarithmic identities.