Problem 29 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(3^{2 x}=80\)

Step-by-Step Solution

Answer

The solution to the exponential equation is \(x \approx 2.022\).

Step 1: Rewrite the Equation Using Logarithms

To approach this problem, first need to rewrite the exponential equation using logarithms. This can be achieved by applying the natural logarithm (because it simplifies calculations), on both sides of the equation. The equation will look like this: \(\ln(3^{2x}) = \ln(80)\).

Step 2: Use the Properties of Logarithms to Simplify the Equation

You can use the property of logarithms \(\ln(a^b) = b \ln(a)\) which allows you to bring the power in front in the equation. In doing so, you get \(2x \ln(3) = \ln(80)\).

Step 3: Solve for x

Now isolate x by dividing through by \(2 \ln(3)\). Therefore, the isolated x is given by the equation \(x = \frac{\ln(80)}{2 \ln(3)}\).

Step 4: Approximate x to Three Decimal Places

Using a scientific calculator, input the equation \(x = \frac{\ln(80)}{2 \ln(3)}\) to find that \(x \approx 2.022\). You should round this to three decimal places, thus \(x \approx 2.022\).

Key Concepts

Logarithms

Understanding logarithms can unlock an entirely new way of approaching exponential equations. A logarithm essentially answers the question: "What power must we raise a base number to obtain another number?" For example, in the equation \(3^{2x} = 80\), we are trying to solve for \(x\), and logarithms help us do just that.
Logarithms can be denoted in different bases, but the natural logarithm, represented by \(\ln\), is particularly useful for solving exponential equations like the one above. They allow us to transform equations in a way that makes them easier to solve. For our problem, by taking the logarithm of both sides, we swap the exponential form for something we can deconstruct more easily.
This process begins an algebraic sequence that leads us to find the value of \(x\). Without logarithms, solving this kind of equation would be significantly more complex.

Solving Algebraically

Solving an equation algebraically involves manipulating the equation step by step while following mathematical rules and properties. When dealing with exponential equations, using logarithms is a typical approach.
Let's break down the steps for solving the given equation \(3^{2x} = 80\) using logarithms:

• Take the natural logarithm of both sides: \(\ln(3^{2x}) = \ln(80)\).
• Apply the power rule of logarithms to simplify: \(2x \ln(3) = \ln(80)\).
• Isolate \(x\): Divide both sides by \(2 \ln(3)\) to get \(x = \frac{\ln(80)}{2 \ln(3)}\).

Each of these steps involves straightforward algebraic manipulation, using known properties. However, the application of these properties requires practice and understanding.

Approximation

Sometimes, exact solutions to equations may not be apparent or necessary. This is where approximation becomes valuable, particularly when dealing with logarithms. In our context, we want to find an approximate value for \(x\) to three decimal places.
Once we have our expression for \(x\) in terms of logarithms, \(x = \frac{\ln(80)}{2 \ln(3)}\), we use a calculator to find numerical values for these logarithms:

• Calculate \(\ln(80)\) and \(\ln(3)\) using a scientific calculator.
• Divide the result of \(\ln(80)\) by \(2\ln(3)\).
• Round the resulting value to three decimal places.

Following these steps yields \(x \approx 2.022\), providing a usable, approximate solution.

Properties of Logarithms

The properties of logarithms are key tools for solving exponential equations algebraically. They allow us to manipulate and simplify logarithmic expressions in powerful ways.
In the given solution, two primary properties of logarithms come into play:

• Power Rule: \(\ln(a^b) = b \ln(a)\). It allows us to move the exponent in front, transforming \(\ln(3^{2x})\) into \(2x \ln(3)\). This step is pivotal in isolating \(x\).
• Logarithm of a Product: Although not explicitly used here, \(\ln(ab) = \ln(a) + \ln(b)\) is another property that can simplify expressions involving products.

Understanding these properties equips you with strategies to tackle various exponential and logarithmic equations effectively. Remembering these properties can make the task of solving equations much more straightforward, transforming seemingly complex problems into manageable ones.