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Chapter 4: Problem 41

Simplify the expression. $$ \log _{2}\left(\frac{1}{16}\right) $$

Short Answer

Expert verified
-4

Step by step solution

01

Rewrite the Fraction as a Power of 2

Recognize that \( \frac{1}{16} \) can be written as \( 2^{-4} \) because \( 16 = 2^4 \).
02

Apply the Logarithm Property

Use the property of logarithms \( \log_b(b^a) = a \) to simplify \( \log_{2}(2^{-4}) \).
03

Simplify the Expression

From the previous property, \( \log_{2}(2^{-4}) = -4 \). Therefore, the expression simplifies to \ -4 \.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Functions
Logarithmic functions play a crucial role in many areas of mathematics, from algebra to calculus. These functions are the inverses of exponential functions. When you see \( log_b(x) \), you are finding the power to which base \( b \) must be raised to yield the number \( x \).
For example, \( log_2(8) = 3 \) because \( 2^3 = 8 \). Understanding this definition is foundational before delving into specific properties.
Note that the base of a logarithm must always be a positive number and cannot be 1. The most common bases are 10 (common logarithm) and \( e \) (natural logarithm), though for various problems, such as the one provided, base 2 might appear often.
Simplification of Logarithms
Simplifying logarithms involves a few key properties. Recognizing these properties allows you to break down complex logarithmic expressions into something manageable.

Key Properties:
  • Log of a Power: The property \( \log_b(b^a) = a \) tells us that the logarithm of a base raised to a power just returns that power.
  • Product Rule: \( log_b(mn) = log_b(m) + log_b(n) \). This property is useful when dealing with multiplications inside the log.
  • Quotient Rule: \( log_b(\frac{m}{n}) = log_b(m) - log_b(n) \) is handy when dealing with divisions inside the log.
  • Power Rule: \( log_b(m^n) = n \cdot log_b(m) \). This rule is convenient when the argument of the log is a power.
These properties can help you tackle various logarithmic problems and simplify expressions.
For the given exercise, we used the Log of a Power property to simplify the expression \( log_2(2^{-4}) = -4 \).
Powers and Exponents
Powers and exponents are foundational to understanding logarithms. In the exercise, we first rewrote the fraction \( \frac{1}{16} \) as \( 2^{-4} \). This is because exponent rules state that \( \frac{1}{a^n} = a^{-n} \).

Understanding Exponents:
  • Positive Exponents: Indicate how many times to multiply the base by itself. For example, \( 2^3 = 2 \times 2 \times 2 \).
  • Negative Exponents: Indicate the reciprocal of the base raised to the corresponding positive power. For instance, \( 2^{-3} = \frac{1}{2^3} \).
  • Zero Exponent: Any base raised to the power of zero equals one: \( a^0 = 1 \).
These exponent rules tie directly to logarithms, as logarithms essentially ask the question, 'Given a base and a number, what exponent produces that number?'
Hence, the problem \( log_2(\frac{1}{16}) \) makes sense when analyzed through the lens of powers and exponents.

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Most popular questions from this chapter

Explain how to use the change-of-base formula and explain why it is important.

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(7^{4 x-1}=3^{5 x}\)

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log _{6}(7 x-2)=1+\log _{6}(x+5)\)

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(10^{5+8 x}+4200=84,000\)

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log (p+17)=4.1\)
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