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Chapter 2: Problem 9

In Exercises \(9-22,\) rewrite the given expression without using any exponentials or logarithms. $$ \log _{5}(1 / 125) $$

Short Answer

Expert verified
The expression simplifies to \(-3\).

Step by step solution

01

Identify the Expression

We need to rewrite the expression \( \log_{5}(1/125) \) without using any exponents or logarithms. The expression involves a logarithm to the base 5.
02

Rewrite the Fraction as a Power

Remember that \(125 = 5^3\). So, \(1/125\) can be rewritten as \((5^3)^{-1}\). By applying the power of a power rule, we have: \(1/125 = 5^{-3}\).
03

Apply Logarithmic Property

Using the property \( \log_b(a^c) = c \cdot \log_b(a) \), we can rewrite the given logarithmic expression: \( \log_{5}(5^{-3}) = -3 \cdot \log_{5}(5)\).
04

Evaluate the Logarithm

Since \( \log_{5}(5) = 1\), we can simplify the expression to: \(-3 \cdot 1 = -3\). Therefore, \( \log_{5}(1/125) = -3\).

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Key Concepts

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

Fraction as a Power
In mathematical expressions, we often need to rewrite fractions as powers to simplify and solve equations involving them. A fraction like \( \frac{1}{125} \) can initially seem complex to work with, especially within logarithmic or exponential expressions. However, recognizing the fraction as a base raised to a negative power can simplify the process.
To understand, consider the fraction \( \frac{1}{125} \). Knowing that 125 can be expressed as \( 5^3 \) makes it easier to reframe the fraction.
  • When we take \( \frac{1}{125} \), we are essentially finding the reciprocal of 125.
  • The reciprocal of a number like \( a^n \) is \( a^{-n} \).
  • Therefore, \( \frac{1}{125} \) becomes \( 5^{-3} \).
This conversion is vital for solving problems involving exponential growth or decay, simplifying expressions, or in this case, evaluating a logarithm.
Logarithmic Properties
Logarithmic properties are powerful tools that help simplify complex logarithmic expressions. One crucial property is the power rule of logarithms, which states that \( \log_b(a^c) = c \cdot \log_b(a) \). This property allows us to take the exponent out of the logarithm, turning a potentially complex equation into something much more manageable.
  • In the expression \( \log_{5}(5^{-3}) \), the power rule allows us to simplify the expression by extracting the exponent, -3.
  • This results in: \( -3 \cdot \log_{5}(5) \).
Furthermore, recognizing simple logarithmic identities can further aid in evaluation. For instance:
  • \( \log_b(b) = 1 \) for any base \( b \), since any non-zero number raised to 1 is itself.
  • Using this identity, \( \log_{5}(5) \) equals 1.
Thus, applying these properties and identities together, we simplify the log expression significantly to a number, in this case, -3.
Exponential Expressions
The concept of exponential expressions often forms the basis of equations in mathematics, capturing growth and decay processes. These expressions feature a constant base raised to a power, such as \( a^b \).
  • They represent repeated multiplication, where an exponent can be positive, negative, or zero.
  • Negative exponents represent the reciprocal, hence \( 5^{-3} \) means \( \frac{1}{5^3} \).
  • Zero exponents denote results equal to 1, since \( a^0 = 1 \), provided \( a eq 0 \).
Mastering these expressions enables students to manipulate equations with ease, essential when dealing with logarithms or simplifying complex expressions. By skillfully managing exponents, one can transform \( 1/125 \) into \( 5^{-3} \), simplifying the entire logarithmic evaluation.

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