Problem 40 - College Algebra

Question

Can the power property of logarithms be derived from the power property of exponents using the equation \(b^{x}=m ?\) If not, explain why. If so, show the derivation.

Step-by-Step Solution

Answer

Yes, the power property of logarithms can be derived using the power property of exponents.

Step 1: Review the Power Property of Exponents

The power property of exponents states that \((b^x)^n = b^{x \cdot n}\). This indicates that raising a number with an exponent to another power results in multiplying the exponents.

Step 2: Review the Power Property of Logarithms

The power property of logarithms states that \(\log_b(m^n) = n \cdot \log_b(m)\). This means that the logarithm of a number with an exponent can be simplified by pulling the exponent in front of the logarithm.

Step 3: Express the Equation in Terms of Logarithms

Given the exponential form \(b^x = m\), take the logarithm on both sides of the equation. This gives us \(\log_b(b^x) = \log_b(m)\).

Step 4: Apply the Power Property of Logarithms

Using the power property of logarithms from Step 2, \(\log_b(b^x)\) can be simplified to \(x \cdot \log_b(b)\). Since \(\log_b(b) = 1\), we have \(x \cdot 1 = x\). Thus, \(x = \log_b(m)\).

Step 5: Reverse the Steps to Demonstrate the Derivation

Starting from \(x = \log_b(m)\), use the power property by considering \(m = b^x\). Thus, if \(m^n = (b^x)^n\), then by the power property of exponents \((b^x)^n = b^{x \cdot n}\). Therefore, using logarithms, \(\log_b(m^n) = \log_b(b^{x \cdot n}) = x \cdot n = n \cdot \log_b(m)\), showing the derivation is consistent.

Key Concepts

Power Property of Exponents

Understanding the **power property of exponents** is essential for tackling both exponential and logarithmic equations. The rule states that when you raise a power to another power, you multiply the exponents. In mathematical terms, this is expressed as \((b^x)^n = b^{x \cdot n}\). Here’s why it works:

• You start with a base \(b\) raised to an exponent \(x\).
• Then, you further raise this expression to another power \(n\).
• The result combines the influences of both exponents by multiplying them together.

Breaking it down helps to see how exponents manage repeated multiplication of a base. If you apply this property to a basic example, like \((2^2)^3\), you multiply 4 (or \(2^2\)) by itself 3 times, giving you \((2^6)\). This fundamental technique is a stepping stone to more complex concepts involving exponents.

Exponential Equations

**Exponential equations** are equations where the variable appears as an exponent. They often take the form \(b^x = m\), where \(b\) is the base, \(x\) is the exponent or power, and \(m\) is the result. Solving them requires converting the format to make the exponent more convenient:

• One way to solve exponential equations involves taking the logarithm of both sides. This can simplify the equation since it allows you to bring the exponent down.
• For example, if you have \(b^x = m\), taking the log base \(b\) gives you \(\log_b(b^x) = \log_b(m)\).
• Using properties of logarithms (since \(\log_b(b) = 1\)), you can simplify it to \(x = \log_b(m)\), extracting \(x\) from the exponent position.

Exponential equations are common in growth and decay problems in sciences like biology and finance, where understanding how quantities change exponentially over time is crucial.

Logarithmic Equations

A **logarithmic equation** involves logarithms. These equations are primarily used to solve for variables when they appear as exponents. Understanding the basics is crucial for effective problem solving:

• The core idea is that a logarithm is the inverse operation of exponentiation, which allows for solving exponential equations easily.
• For instance, consider the formula \(x = \log_b(m)\), which tells us \(b^x = m\).
• Logarithmic equations often switch the roles of knowns and unknowns compared to exponential forms.

To solve logarithmic equations, one usually rewrites them in exponential form or uses logarithmic properties such as the power property \(\log_b(m^n) = n \cdot \log_b(m)\). This property shows how you can move an exponent out front, which simplifies the solving process. This is particularly useful when tackling more complex equations where logarithms provide a simpler framework for solution derivation.