Problem 29
Question
For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation. $$ \log \left(\frac{1}{100}\right)=-2 $$
Step-by-Step Solution
Verified Answer
The exponential equation is \( 10^{-2} = \frac{1}{100} \).
1Step 1: Understand the Logarithm Definition
The definition of a logarithm states that if \( \log_b(a) = c \), then this is equivalent to saying \( b^c = a \). This means that the base \( b \) raised to the power of \( c \) equals \( a \).
2Step 2: Identify Components of the Given Logarithmic Equation
In the equation \( \log \left(\frac{1}{100}\right) = -2 \), the base is not explicitly written, which means it is the common logarithm base of 10. Hence, we identify: \( b = 10 \), \( a = \frac{1}{100} \), and \( c = -2 \).
3Step 3: Use the Definition to Rewrite as an Exponential Equation
Using the definition from Step 1, substitute the components identified in Step 2: \( 10^{-2} = \frac{1}{100} \). This expresses the original logarithmic equation as an exponential equation.
Key Concepts
Exponential EquationsCommon Logarithm BaseLogarithm Definition
Exponential Equations
Exponential equations are equations in which variables appear as exponents. These types of equations might look tricky at first, but they can become more understandable if we decompose them step by step.
Typically, an exponential equation looks like this: \(b^c = a\). Here:
Typically, an exponential equation looks like this: \(b^c = a\). Here:
- \(b\) is the base
- \(c\) is the exponent (or the power)
- \(a\) is the result.
- To transform a logarithm into an exponential equation, we identify its components and apply the definition methodically.
- Use the format \(b^c = a\) to clearly see the relationship between every part of the equation.
- This understanding makes solving real-world problems involving exponential growth or decay much clearer.
Common Logarithm Base
The common logarithm is one that uses base 10, often denoted simply as \(\log\). When no base is written in a logarithmic equation, it's typically the common logarithm.
In our equation, \(\log \left(\frac{1}{100}\right) = -2\), the base is not specified, which tells us it's base 10 by default. This is a handy convention because it means we don't have to specify it every time, simplifying both writing and understanding logarithmic expressions.
In our equation, \(\log \left(\frac{1}{100}\right) = -2\), the base is not specified, which tells us it's base 10 by default. This is a handy convention because it means we don't have to specify it every time, simplifying both writing and understanding logarithmic expressions.
- The number 10 is used as a base because it corresponds nicely with the decimal system, the foundation of most numeric systems we use daily.
- When converting logarithmic equations which don't specify a base, always assume it's base 10 unless otherwise noted.
- Familiarity with base 10 logs is critical not just in mathematical contexts, but also in fields like science and engineering where calculations frequently hinge on powers of 10.
Logarithm Definition
Understanding logarithms starts with recognizing how they connect exponential and multiplicative inverses. The definition is key: Given \(\log_b(a) = c\), it tells us that the base \(b\) raised to the power \(c\) equals \(a\).
This means that logarithms help us reverse the process of exponentiation by answering the question: "To what power must the base be raised to obtain this number?"
This means that logarithms help us reverse the process of exponentiation by answering the question: "To what power must the base be raised to obtain this number?"
- It breaks down the process of finding powers into more manageable steps.
- In practice, this means we can take complex multiplications and convert the calculations into simpler additions by working with logarithms.
- For the equation \(\log \left(\frac{1}{100}\right) = -2\), we determined that \(b\), the base, is 10. The solution such as \(10^{-2} = \frac{1}{100}\) emerges from this understanding.
Other exercises in this chapter
Problem 29
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