Problem 8 - Calculus of a Single Variable

Question

Exponential and Logarithmic Forms of Equations In Exercises \(5-8,\) write the exponential equation as a logarithmic equation or vice versa. $$ \begin{array}{l}{\text { (a) } \log _{3} \frac{1}{9}=-2} \\ {\text { (b) } 49^{1 / 2}=7}\end{array} $$

Step-by-Step Solution

Answer

The exponential form of the logarithmic equation \(\log_{3}\frac{1}{9} = -2\) is \(3^{-2}=\frac{1}{9}\). The logarithmic form of the exponential equation \(49^{1/2} = 7\) is \(\log_{49}7 = \frac{1}{2}\).

Step 1: Conversion of Logarithmic Equation into Exponential Form

For part (a), the given equation is in logarithmic form which is \(\log_{3}\frac{1}{9} = -2\). To convert it into exponential form, apply the formula \(\log_{a}c=b\) is equivalent to \(a^{b}=c\). Therefore, the exponential form of the equation is \(3^{-2}=\frac{1}{9}\).

Step 2: Conversion of Exponential Equation into Logarithmic Form

For part (b), the given equation is in exponential form, which is \(49^{1/2} = 7\). To convert it into logarithmic form, apply the formula \(a^{b}=c\) is equivalent to \(\log_{a}c=b\). Therefore, the logarithmic form of the equation is \(\log_{49}7 = \frac{1}{2}\).

Key Concepts

Logarithmic Form

A logarithmic equation is an expression that involves a logarithm. The logarithm is a mathematical operation that determines the exponent or power to which a base number must be raised to produce a given number. It often is expressed using the syntax:

• \( \log_{a}(c) = b \) - This means that "\( a \) raised to the power of \( b \) equals \( c \)."

For example, the equation \( \log_{3} \frac{1}{9} = -2 \) is in logarithmic form. Here, the base is 3, the exponent that 3 must be raised to is

• -2, and the result is \( \frac{1}{9} \).

Understanding the components and structure of a logarithmic expression helps convert it easily to an exponential equation, which often simplifies calculations and reveals patterns that are not plainly visible in logarithmic form. The process follows from the definition: just identifying the base, the answer, and the number will guide your conversion.

Exponential Form

An exponential equation reveals the relationship between numbers, where a constant base is raised to a power to yield a result. It is generally represented as:

• \( a^{b} = c \) - Where \( a \) is the base, \( b \) is the exponent, and \( c \) is the result.

In the example \( 49^{1/2} = 7 \), the base is 49, the exponent is \( 1/2 \), and the result is 7.

• The way exponential expressions show the components of an equation can make it easier to solve problems by highlighting the base and the result.

This form emphasizes the power and its endpoint, prompting a clear understanding of multiplying the base a certain number of times (or root extraction in fractional powers). A strong grasp of exponential expressions is essential in mathematics, not only for solving equations but also for understanding growth patterns, decay, and various scientific phenomena.

Logarithmic and Exponential Formula Conversion

Conversion between logarithmic and exponential forms is a key skill in algebra. This conversion uses a fundamental understanding that both forms are two sides of a mathematical coin. The transformations are represented by the following general formulas:

• From logarithmic to exponential: \( \log_{a}(c) = b \) converts to \( a^{b} = c \).
• From exponential to logarithmic: \( a^{b} = c \) converts to \( \log_{a}(c) = b \).

Learning these conversions is important because it allows flexibility in solving equations. Sometimes, an equation is easier to solve or analyze in its exponential form, while other times, the logarithmic form is more practical. For instance, changing \( \log_{3} \frac{1}{9} = -2 \) into its exponential form \( 3^{-2} = \frac{1}{9} \) allows us to directly see and check the relationship between numbers. Similarly, converting \( 49^{1/2} = 7 \) into \( \log_{49} 7 = \frac{1}{2} \) provides a perspective where exponential relationships can be understood in terms of logarithms, offering insight into the powers involved. Practicing these conversions helps enhance problem-solving skills across different branches of mathematics.