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Ch. 4 - Exponential and Logarithmic Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 5, Problem 105

In Exercises 105–108, evaluate each expression without using a calculator. log5 (log7 7)

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1
Recognize that the expression is \( \log_5 (\log_7 7) \), which means the logarithm base 5 of the logarithm base 7 of 7.
Evaluate the inner logarithm first: \( \log_7 7 \). Recall that \( \log_b b = 1 \) for any base \( b > 0 \) and \( b \neq 1 \).
Since \( \log_7 7 = 1 \), substitute this value back into the original expression to get \( \log_5 1 \).
Recall that \( \log_b 1 = 0 \) for any valid base \( b \), because \( b^0 = 1 \).
Therefore, the expression simplifies to \( \log_5 1 = 0 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Logarithm Definition

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. Understanding this definition helps in interpreting and simplifying logarithmic expressions.
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Logarithm of the Base Itself

The logarithm of a base raised to itself, such as log_b(b), always equals 1 because b^1 = b. This property simplifies expressions like log7(7) to 1, which is crucial for evaluating nested logarithms.
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Evaluating Nested Logarithms

Nested logarithms involve one logarithm inside another, like log5(log7(7)). To evaluate, simplify the inner logarithm first, then apply the outer logarithm. This stepwise approach avoids calculator use and clarifies the expression.
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