Problem 12 - Algebra for College Students

Question

Solve each of the equations. $$ \left(\frac{2}{3}\right)^{n}=\frac{9}{4} $$

Step-by-Step Solution

Answer

The solution is \( n = -3.713 \).

Step 1: Understand the Problem

The equation to solve is \( \left(\frac{2}{3}\right)^{n} = \frac{9}{4} \). We need to find the value of \( n \) that satisfies this equation.

Step 2: Use Logarithms to Solve for n

To solve for \( n \), take the natural logarithm of both sides of the equation: \[ \ln\left(\left(\frac{2}{3}\right)^{n}\right) = \ln\left(\frac{9}{4}\right) \] Using the power rule of logarithms, \( \ln\left(a^b\right) = b \ln(a) \), we get: \[ n \cdot \ln\left(\frac{2}{3}\right) = \ln\left(\frac{9}{4}\right) \]

Step 3: Isolate the Variable n

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

Step 4: Calculate the Value of n

Calculate \( \ln\left(\frac{9}{4}\right) \) and \( \ln\left(\frac{2}{3}\right) \) using a calculator. Then compute:\[ n = \frac{\ln\left(\frac{9}{4}\right)}{\ln\left(\frac{2}{3}\right)} \approx \frac{1.504}{-0.405} \approx -3.713 \]

Step 5: Verify the Solution

To ensure the solution is correct, substitute \( n = -3.713 \) back into the original equation and check if both sides are equal. Calculate \( \left(\frac{2}{3}\right)^{-3.713} \), and verify it is approximately \( \frac{9}{4} \).

Key Concepts

Logarithms in Algebra

Logarithms are incredibly useful tools in algebra, especially when dealing with exponential equations. They help us "re-write" equations containing exponents in a form that's easier to manipulate. This process involves the inverse of exponentiation.
To put it simply, if you have an equation like the one in our exercise, taking the logarithm of both sides allows us to transform the exponential expression into a linear one. This is because logarithms help bring the exponent "down" in front of the log expression due to the properties of logarithms.
In algebra, common logarithms (base-10) and natural logarithms (base-

• e, approximately 2.718) are often used to solve equations. Natural logarithms, represented as \( \ln \), are particularly useful in calculus and advanced algebra.

Understanding how to apply logarithms in algebra can simplify complex equations and is an essential skill when solving various types of mathematical problems.

Solving Equations

To solve equations effectively, especially those involving exponents or logarithms, you need a clear strategy. In our specific problem, the goal was to find the value of \( n \) that satisfies \( \left(\frac{2}{3}\right)^{n} = \frac{9}{4} \).
We started by taking the natural logarithm of both sides. This step reduced the complexity of the exponent by using logarithmic properties. After applying the logarithm, the equation became linear: \( n \cdot \ln\left(\frac{2}{3}\right) = \ln\left(\frac{9}{4}\right) \).
Isolating the variable \( n \) was key. To do this, we divided both sides by \( \ln\left(\frac{2}{3}\right) \). This allowed us to directly solve for \( n \) using a simple arithmetic division. Solving such equations usually involves:

• Identifying the form of the equation (exponential, quadratic, etc.).
• Selecting the appropriate mathematical techniques (in this case, logarithms).
• Applying the chosen techniques step-by-step while checking your work.
• Interpreting the result in the context of the equation.

Power Rule of Logarithms

The power rule of logarithms is a fundamental property that makes logarithms so useful, especially in transforming equations where exponents are involved. It states that \( \ln\left(a^b\right) = b \cdot \ln(a) \), which effectively "brings down" the exponent as a coefficient.
In mathematical problems involving exponential expressions, using the power rule can simplify equations from exponential to linear forms. This transformation was crucial in solving our exercise.
Here's what this rule allows:

• Turning any problem of the form \( a^b \) into a multiplicative problem \( b \cdot \ln(a) \). By doing so, it becomes easier to solve for unknown variables.
• This property is versatile and applies to equations involving exponent roots, powers, and any form of exponential expressions.

Utilizing the power rule effectively allows us to tackle complex equations methodically, making it an indispensable tool in algebra and beyond.