Problem 19 - Calculus of a Single Variable

Question

Solving an Equation In Exercises 19-24, solve for \(x\) or \(b\) . $$ \begin{array}{l}{\text { (a) } \log _{10} 1000=x} \\ {\text { (b) } \log _{10} 0.1=x}\end{array} $$

Step-by-Step Solution

Answer

The solution to the equations \( \log_{10} 1000=x \) and \( \log_{10} 0.1=x \) are \( x = 3 \) and \( x = -1 \) respectively.

Step 1: Solve for equation (a)

To solve \( \log_{10} 1000=x \), we first need to express the logarithm in exponential form, which gives us \( 10^x = 1000 \). We clearly see that \( 10^3 = 1000 \), hence \( x = 3 \).

Step 2: Solve for equation (b)

Next, to solve \( \log_{10} 0.1=x \), we express this in exponential form, which yields \( 10^x = 0.1 \). Rewriting 0.1 as a power of 10 gives us \( 10^x = 10^{-1} \). Therefore, \( x = -1 \).

Key Concepts

Exponential Functions

Exponential functions are all about understanding growth and decay. They involve expressions where a constant base is raised to a variable exponent, like in the equation \( y = a^x \). Here, \( a \) represents the base, and \( x \) is the power to which the base is raised.​
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When dealing with exponential functions, it’s crucial to recognize the role of the base and the exponent. The base determines the function’s rate of growth or decay. If the base is greater than one, the function shows exponential growth, meaning it increases rapidly. For example, if \( a = 2 \), the function \( y = 2^x \) doubles for every increase in \( x \).
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In contrast, if the base is between 0 and 1, the function demonstrates exponential decay, like in \( y = 0.5^x \), which halves with every increase in \( x \). When solving exponential equations, converting them into logarithmic form can make finding the value of the variable much easier. In the exercise provided, converting the logarithmic equations into exponential form was a key step in solving for \( x \).

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. There are various strategies to solve equations, and one common approach involves isolating the variable.
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Let's break down solving the equations in the exercise:

• **Equation (a):** \( \log_{10} 1000 = x \). First, convert the logarithmic equation into exponential form, giving \( 10^x = 1000 \). Recognize that \( 10^3 = 1000 \), hence \( x = 3 \).
• **Equation (b):** \( \log_{10} 0.1 = x \). Again, convert into exponential form: \( 10^x = 0.1 \). Express \( 0.1 \) as \( 10^{-1} \), so \( x = -1 \).

For each equation, understanding how to express numbers as powers of 10 is crucial, allowing us to equate the exponents and solve the equation effectively.
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This process shows the direct link between exponential functions and logarithmic equations, vastly simplifying the solving process when you recognize how to switch forms.

Base 10 Logarithms

Base 10 logarithms, or common logarithms, use 10 as the base of the logarithm. They are often represented as \( \log \) without the base explicitly stated, as it is typically understood to be 10.
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Logarithms are essentially the inverse operations of exponentiation. They answer the question: "To what power must the base be raised, to obtain a certain number?"
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For instance, in \( \log_{10} 1000 \), we’re asking: "10 raised to what power gives us 1000?" The answer, as we calculated, is 3 since \( 10^3 = 1000 \). Similarly, \( \log_{10} 0.1 \) asks for the power of 10 that equals 0.1, which is \(-1\) because \( 10^{-1} = 0.1 \).
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Base 10 logarithms are particularly useful when dealing with scientific calculations and measuring quantities like pH in chemistry or decibels in acoustics, as they allow us to easily work with very large or very small numbers by reducing them to more manageable powers of ten. When encountering logarithmic equations, translating them to exponential form helps in visualizing and solving them effectively.