Problem 5 - 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. \(\left(\frac{4}{25}\right)^{-1 / 2}=\frac{5}{2}\)

Step-by-Step Solution

Answer

The exponential equation \( \left( \frac{4}{25} \right)^{-1/2} = \frac{5}{2} \) can be rewritten as the logarithmic equation \( \log_{\left(\frac{4}{25}\right)}\left(\frac{5}{2}\right) = -\frac{1}{2} \).

Step 1: Understanding the Given Equation

We are given the exponential equation \( \left( \frac{4}{25} \right)^{-1/2} = \frac{5}{2} \). This equation is in the form of \( b^a = c \), where \( b = \frac{4}{25} \), \( a = -\frac{1}{2} \), and \( c = \frac{5}{2} \).

Step 2: Applying the Logarithmic Form

Using the property \( b^a = c \) if and only if \( \log_b(c) = a \), we need to convert this exponential equation into a logarithmic equation. We set \( b = \frac{4}{25} \), \( c = \frac{5}{2} \), and \( a = -\frac{1}{2} \).

Step 3: Write the Logarithmic Equation

According to the property, \( \log_b(c) = a \) becomes \( \log_{\left(\frac{4}{25}\right)}\left(\frac{5}{2}\right) = -\frac{1}{2} \). This is the logarithmic form of the given exponential equation.

Key Concepts

logarithmic form

The logarithmic form of an equation is a way of expressing exponential equations using logarithms. Logarithms and exponents are closely related since a logarithm effectively tells us the power we need to raise a base to obtain a certain number. For the given exercise, we had an exponential equation:

• \( \left(\frac{4}{25}\right)^{-1/2} = \frac{5}{2} \)

Here, we converted this form to its logarithmic equivalent using the definition that if \( b^a = c \) then \( \log _b(c) = a \).

• The base \( b \) is \( \frac{4}{25} \)
• The solution \( c \) is \( \frac{5}{2} \)
• The exponent \( a \) is \( -\frac{1}{2} \)

Thus, the logarithmic form of this equation is:

• \( \log_{\left(\frac{4}{25}\right)}\left(\frac{5}{2}\right) = -\frac{1}{2} \)

This transformation highlights that all exponential equations can be represented as logarithms, providing an alternate perspective on the relationships between the base, exponent, and result.

exponential form

Exponential form is a mathematical expression where a constant (known as the base) is raised to a certain power (or exponent) to yield another quantity. It's a concise way of expressing repeated multiplication. Let's look at the exponential equation from the exercise:

• \( \left(\frac{4}{25}\right)^{-1/2} = \frac{5}{2} \)

In this form,

• The base is \( \frac{4}{25} \)
• The exponent is \(-\frac{1}{2} \)
• The result is \( \frac{5}{2} \)

Converting logarithmic expressions to exponential forms makes it easier to understand exponential growth or decay, comparing quantities, and solving equations involving powers.
Examples of rewriting equations help improve understanding:

• If \( \log_2(8) = 3 \), the equivalent exponential form is \( 2^3 = 8 \).

Understanding exponential forms allows you to rewrite the expressions and equations effectively, assisting in solving and simplifying mathematical problems.

Theorem 6.2

Theorem 6.2 is a fundamental principle linking exponential and logarithmic equations. It states that if \( b^a = c \), then \( \log_b(c) = a \). This theorem serves a key role in transforming these two forms into each other, enhancing our ability to solve both types of equations easily.
The application of this theorem:

• Allows us to convert exponential equations into logarithmic equations, giving an alternative method to study such mathematical concepts.
• Provides insight into logarithmic behavior, which directly contrasts with exponential functions.

By leveraging this theorem, we can solve an exponential or logarithmic equation by simply changing its form, as shown in our exercise. Here:

• Exponential: \( \left(\frac{4}{25}\right)^{-1/2} = \frac{5}{2} \)
• Logarithmic: \( \log_{\left(\frac{4}{25}\right)}\left(\frac{5}{2}\right) = -\frac{1}{2} \)

Understanding Theorem 6.2 enriches your capability to manipulate and interpret these two expressions, fostering a deeper understanding of their mathematical relationship.