Problem 20 - Intermediate Algebra

Question

For Problems \(1-34\), solve each equation. $$ \left(\frac{2}{3}\right)^{x}=\frac{9}{4} $$

Step-by-Step Solution

Answer

The solution is \( x \approx -2 \).

Step 1: Understand the Equation

We are given the equation \( \left( \frac{2}{3} \right)^x = \frac{9}{4} \). Our objective is to find the value of \( x \).

Step 2: Take Logarithm

To solve for \( x \) when it's an exponent, it's helpful to use logarithms. Take the natural logarithm of both sides: \( \ln \left( \left( \frac{2}{3} \right)^x \right) = \ln \left( \frac{9}{4} \right) \).

Step 3: Apply Logarithm Power Rule

Using the power rule for logarithms, \( \ln(a^b) = b \ln(a) \), we have \( x \ln \left( \frac{2}{3} \right) = \ln \left( \frac{9}{4} \right) \).

Step 4: Solve for x

To isolate \( x \), divide both sides by \( \ln \left( \frac{2}{3} \right) \): \( x = \frac{\ln \left( \frac{9}{4} \right)}{\ln \left( \frac{2}{3} \right)} \).

Step 5: Calculate the Values

Use a calculator to find the values: \( \ln \left( \frac{9}{4} \right) \approx 0.8109 \) and \( \ln \left( \frac{2}{3} \right) \approx -0.4055 \). Thus, \( x \approx \frac{0.8109}{-0.4055} \approx -2. \)

Key Concepts

Logarithms

Logarithms are incredibly useful in solving equations where the unknown variable is an exponent. They are the inverse operations of exponentiation, meaning that they "undo" exponential functions. For any positive number \(a\), where \(a \, eq 1\), the logarithm base \(a\) of a number is defined as the exponent to which \(a\) must be raised to obtain that number.

In the context of exponential equations, like \( \left( \frac{2}{3} \right)^x = \frac{9}{4} \), logarithms allow us to pull down the exponent, \(x\), making it easier to isolate and solve for. This is why we take the logarithm of both sides of the equation: \( \ln \left( \left( \frac{2}{3} \right)^x \right) = \ln \left( \frac{9}{4} \right) \).

Using natural logarithms (denoted as \(\ln\)) is common due to their properties and ease of use in calculus. However, any logarithmic base could be applied if consistent across the equation. When working with logarithms, it’s crucial to remember the basic laws, such as the logarithm power rule, which makes them powerful tools in algebra and calculus.

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. In our exercise, the goal is to find \(x\).

The equation \( \left( \frac{2}{3} \right)^x = \frac{9}{4} \) initially presents a challenge because \(x\) is in the exponent. This is where the logarithm comes into play. By taking the natural logarithm of both sides, we change the equation from an exponential one to a linear one.

Steps to solving such equations typically include:

• Identifying the type of equation and simplification needed,
• Applying arithmetic operations to maintain equilibrium in the equation,
• Using algebraic rules, such as logarithms, to transform and simplify,
• Reversing operations to isolate the variable.

These systematic steps enhance not only solving capability but empower solving equations of various complexities.

Algebraic Manipulation

Algebraic manipulation is crucial in solving equations and simplifying expressions. It involves rearranging and simplifying the parts of equations to isolate the desired variable. This exercise demonstrates several key techniques.

First, after applying logarithms, the equation \( \ln \left( \left( \frac{2}{3} \right)^x \right) = \ln \left( \frac{9}{4} \right) \) transforms using the power rule of logarithms, leading to \( x \ln \left( \frac{2}{3} \right) = \ln \left( \frac{9}{4} \right) \).

Next, continue using algebraic manipulation by solving for \(x\). This involves rearranging to isolate \(x\) on one side, by dividing both sides by \( \ln \left( \frac{2}{3} \right) \). This step showcases the essential skill of manipulating equations to make the unknown variable subject to the equation.

Finally, the calculation and simplification of these expressions with a calculator ensure that you arrive at the numerical solution of \(x \approx -2\). With practice, such skills in algebraic manipulation improve significantly, allowing for tackling more complex mathematical problems efficiently.