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Solve each equation.
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Ch. 4 - Exponential and Logarithmic Functions
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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TOC.tableOfContents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Capitolo 5, Problema 106
In Exercises 105–108, evaluate each expression without using a calculator. log5 (log2 32)
Guida verificata passo dopo passo1
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.
Evaluate the inner logarithm \( \log_2 32 \). Recall that \( \log_b a = c \) means \( b^c = a \). So, find the exponent \( c \) such that \( 2^c = 32 \).
Since \( 32 = 2^5 \), it follows that \( \log_2 32 = 5 \). Now substitute this value back into the original expression to get \( \log_5 5 \).
Evaluate the outer logarithm \( \log_5 5 \). Using the definition of logarithms, \( \log_b b = 1 \) for any base \( b > 0 \) and \( b \neq 1 \).
Therefore, \( \log_5 5 = 1 \). This completes the evaluation of the original expression.
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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.
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7:30
Logarithms Introduction
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.
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5:14Evaluate Logarithms
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.
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5:36Change of Base Property
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