Problem 320 - Precalculus

Question

Use the definition of a logarithm to rewrite the equation as an exponential equation. \(\log _{324}(18)=\frac{1}{2}\)

Step-by-Step Solution

Answer

Exponential form: \( 324^{\frac{1}{2}} = 18 \).

Step 1: Understanding the Logarithm

Before rewriting the equation, recall the key property of a logarithm: If \( \log_b(a) = c \), then the corresponding exponential form is \( b^c = a \).

Step 2: Identifying Base, Result, and Exponent

In the given logarithmic equation \( \log_{324}(18) = \frac{1}{2} \), identify the base \(b\) as 324, the result of the logarithm \(a\) as 18, and the exponent \(c\) as \(\frac{1}{2}\).

Step 3: Rewriting as an Exponential Equation

Using the property from Step 1, rewrite the given logarithm as the exponential equation: \( 324^{\frac{1}{2}} = 18 \).

Step 4: Verifying the Exponential Form

Calculate \( 324^{\frac{1}{2}} \), which is the square root of 324. Since \( \sqrt{324} = 18 \), the exponential form is consistent with the original equation.

Key Concepts

Exponential Equations

Exponential equations involve expressions where variables appear in exponents. They can be tricky at first, but are powerful tools in math. Let's delve deeper into understanding these unique equations.
If we have an exponential equation of the form \( b^c = a \), like in our exercise, we're essentially saying that the base \( b \) raised to the power of \( c \) results in \( a \). This is a fundamental principle of exponents.
Consider the exponential equation \( 324^{\frac{1}{2}} = 18 \). Here, 324 is the base, \( \frac{1}{2} \) is the exponent, and 18 is the result after performing the operation.
This equation shows a direct conversion from a logarithmic form to its exponential counterpart. By computing \( 324^{\frac{1}{2}} \), which equates to finding the square root of 324, we effectively verify that the resulting value is indeed 18.
Learning to maneuver between logarithmic and exponential forms can greatly improve your ability to solve diverse mathematical problems involving exponents. This skill is vital in tackling more complex algebraic functions and calculus problems, making exponential equations an essential concept to master.

Change of Base

The "Change of Base" formula is an incredible tool that allows us to compute logarithms with ease, regardless of the base. The formula is particularly useful when your calculator only supports common logarithm (base 10) or natural logarithm (base \( e \)).
Typically, the change of base formula states:

• \( \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \)

where \( k \) could be any base of your choice. Most often, \( k \) is chosen to be 10 (common logarithm) or \( e \) (natural logarithm), as calculators generally provide buttons for these.
This switch allows us from rewriting something like \( \log_{324}(18) \) in a different and perhaps more comfortable base, though our given exercise doesn't require a change of base to solve it.
The change of base formula becomes handy in complex calculations where converting to a familiar base can simplify our understanding and operational execution.

Logarithm Properties

Logarithm properties are the rules that govern operations involving logs and they are key to simplifying and solving logarithmic equations. Understanding these properties can demystify many logarithmic operations. Here are some essential properties:

• **Product Property**: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• **Quotient Property**: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• **Power Property**: \( \log_b(M^p) = p \cdot \log_b(M) \)

These properties show how logarithms can transform multiplicative operations into additive ones, thus simplifying complex calculations.
In the exercise, recognizing the relationship \( \log_b(a) = c \) as \( b^c = a \) leverages the very definition of a logarithm. This property is a crucial aspect of understanding how logs convert to exponential forms.
Grasping these fundamental properties allows you to manipulate and simplify logarithmic expressions and tackle exponential equations effectively, broadening your analytical toolkit in mathematics.