Problem 77 - A Graphical Approach to College Algebra

Question

$$\text { Solve each formula for the indicated variable.}$$ $$\log A=\log B-C \log x, \text { for } A$$

Step-by-Step Solution

Answer

\( A = \frac{B}{x^C} \)

Step 1: Understand the Given Equation

The given equation is \( \log A = \log B - C \log x \). We need to solve it for the variable \( A \).

Step 2: Isolate \( \log A \) on One Side

The equation already expresses \( \log A \) in terms of other variables: \( \log A = \log B - C \log x \).

Step 3: Use Properties of Logarithms

Recall that the equation \( \log a - \log b = \log \left(\frac{a}{b}\right) \). Therefore, the equation \( \log A = \log B - C \log x \) can be rewritten as \( \log A = \log \left(\frac{B}{x^C}\right) \).

Step 4: Solve for \( A \)

To solve for \( A \), remove the logarithm by taking the antilogarithm: \( A = \frac{B}{x^C} \).

Step 5: Conclusion: Verified Solution

The equation has been correctly manipulated to solve for \( A \). The solution satisfies the equation, and all steps follow logical mathematical transformations.

Key Concepts

Properties of Logarithms

Logarithms have specific properties that make them very powerful tools in mathematics. One property is the "Product Formula," which states that the logarithm of a product is the sum of the logarithms:

\( \log(a \cdot b) = \log a + \log b \)

Another important property is the "Quotient Formula," allowing us to express the logarithm of a quotient as the difference of two logarithms:

\( \log\left(\frac{a}{b}\right) = \log a - \log b \)

This formula is particularly useful in this exercise because it helps simplify equations by breaking them into manageable parts. Additionally, we have the "Power Rule," where the logarithm of a power is the exponent times the logarithm of the base:

\( \log(a^n) = n \cdot \log a \)

These properties can transform and solve complex logarithmic equations by simplifying and isolating terms. Understanding and applying these rules is essential when working with logarithmic equations.

Isolation of Variables

In algebra, isolating a variable means rearranging an equation so that a specific variable is alone on one side. This is crucial for solving equations. In the context of logarithms, it often involves using properties of logarithms to rewrite terms in a way that isolates the desired variable.
For instance, in our original exercise, we wanted to solve for \( A \). We began with the equation \( \log A = \log B - C \log x \).
By using the "Quotient Formula," the equation transforms into \( \log A = \log\left(\frac{B}{x^C}\right) \), effectively isolating \( \log A \) in terms of other known variables.
This simplification step is critical because it allows for subsequent transformations, such as removing the logarithm by exponentiating both sides, to eventually solve for \( A \). In general, learning how to isolate variables can help tackle a wide array of algebraic problems.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They have the form \( f(x) = b^x \), where \( b \) is the base and \( x \) is the exponent.
These functions play a key role when it comes to inverse operations of logarithms. In our exercise, after isolating \( \log A \), the next step was to solve for \( A \). By remembering that the antilogarithm of \( \log b = c \) results in \( b = 10^c \) (assuming base 10), we can solve \( \log A = \log\left(\frac{B}{x^C}\right) \) by using the exponential function:

\( A = \frac{B}{x^C} \)

Exponential functions allow us to solve equations that involve logarithms by 'undoing' the logarithmic operation. Mastering how they work equips you to handle various equations spanning from simple algebraic to more complex logarithmic expressions.