Problem 11 - Precalculus

Question

Use the property: \(b^{a}=c\) if and only if \(\log _{b}(c)=a\) from Theorem 6.2 to rewrite the given equation in the other form. That is, rewrite the exponential equations as logarithmic equations and rewrite the logarithmic equations as exponential equations. \(\log _{\frac{4}{3}}\left(\frac{3}{4}\right)=-1\)

Step-by-Step Solution

Answer

The equation in exponential form is \(\left(\frac{4}{3}\right)^{-1} = \frac{3}{4}\).

Step 1: Identify the form of the equation

The given equation is in logarithmic form: \(\log_{\frac{4}{3}}\left(\frac{3}{4}\right)=-1\).

Step 2: Apply the theorem

According to Theorem 6.2, \(b^{a}=c\) if and only if \(\log_{b}(c)=a\). Here, \(b = \frac{4}{3}\), \(c = \frac{3}{4}\), and \(a = -1\).

Step 3: Rewrite using exponential form

Apply the theorem to rewrite the logarithmic equation as an exponential equation: \(\left(\frac{4}{3}\right)^{-1} = \frac{3}{4}\).

Step 4: Verify the rewritten equation

Evaluate the exponential equation: \(\left(\frac{4}{3}\right)^{-1}\) is the reciprocal of \(\frac{4}{3}\), which is \(\frac{3}{4}\). The exponential equation is correct.

Key Concepts

Logarithmic Form

In mathematics, a logarithmic equation is an equation that involves the logarithm of an expression. The logarithm is the inverse function to exponentiation and gives the exponent as its output. For instance, if you have

• \( \log_b(c) = a \)

This means the power to which the base \( b \) must be raised to produce the value \( c \) is \( a \). Logarithms help transform multiplicative relationships into additive ones, making calculations and problem-solving easier in many cases.
When working with logarithmic equations, it's crucial to understand that moving from a logarithmic form to an exponential form can simplify solving some types of equations. In our given problem, the expression \( \log_{\frac{4}{3}}\left(\frac{3}{4}\right)=-1 \) tells us that the base \( \frac{4}{3} \) raised to the power of \( -1 \) equals \( \frac{3}{4} \). The process of identifying and using these forms is guided by specific mathematical rules and theorems that we will explore in the following sections.

Exponential Form

Understanding the exponential form of an equation is an essential step in transitioning from logarithmic to exponential relationships, as we just witnessed with the equation \( \log_{\frac{4}{3}}\left(\frac{3}{4}\right)=-1 \). In exponential form, this translates to:

• \( \left(\frac{4}{3}\right)^{-1} = \frac{3}{4} \)

But what does this mean in simple terms? If we raise the base \( \frac{4}{3} \) to the power of \( -1 \), it is the same as taking its reciprocal. Thus, \( \left(\frac{4}{3}\right)^{-1} \) becomes \( \frac{3}{4} \), precisely matching the equation.
Exponential equations are expressions where the unknown variable appears in the exponent. They are typically rewritten as logarithms to solve them more effectively. Going back and forth between these forms ensures complete understanding and provides multiple paths to solve problems. This flexibility in choosing approaches can simplify many complex mathematical scenarios.

Theorem Application

Theorem application is fundamental in converting equations from one form to another. Here, in the problem at hand, Theorem 6.2 plays a vital role. This theorem states that "\( b^a = c \) if and only if \( \log_b(c) = a \)." It acts as a bridge, allowing us to switch seamlessly between logarithmic and exponential notations.

• Understanding theorem applications involve knowing:
  
  • The roles of each component: \( b \), \( a \), and \( c \).
  • How to interpret what these mean in practical problem contexts.
  • And verifying the rewritten equation to ensure it holds true.

In the given context, identifying that \( b = \frac{4}{3} \), \( c = \frac{3}{4} \), and \( a = -1 \), allows us to apply the theorem effectively to rewrite the equation. Confirming both the progression and solution, theorem application verifies that moving from logarithmic to exponential forms (or vice versa) doesn't alter the inherent truth of the equation. This dual-use ability provides elegance and power in mathematical problem solving.