Problem 22 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$250 e^{0.05 x}=400$$

Step-by-Step Solution

Answer

To obtain the numerical value of \(x\), use a calculator to evaluate \(\frac{\ln(1.6)}{0.05}\), and round to three decimal places if needed. So, \(x\) is approximately 4.382.

Step 1: Isolate the Exponential Term

First, isolate the exponential term on one side of the equation. We do that by dividing both sides of the equation by 250 in order to make the equation simpler to handle:\[ e^{0.05 x} = \frac{400}{250} = 1.6 \]

Step 2: Take Natural Logarithm of Both Sides

Apply the natural logarithm to both sides of the equation. Make sure to apply it to the entire side of the equation:\[ \ln(e^{0.05 x}) = \ln(1.6) \]

Step 3: Apply Logarithm Property

When taking the natural logarithm of an exponential the base with its logarithm will cancel out according to the logarithmic identity \(\ln(e^x) = x\), simplifying the left side of the equation:\[ 0.05x = \ln(1.6) \]

Step 4: Solve for x

To isolate \(x\), we can divide both sides of the equation by 0.05:\[ x = \frac{\ln(1.6)}{0.05} \]

Key Concepts

Natural Logarithms

The natural logarithm, denoted as \( \ln \), is a logarithm with the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828.

Natural logarithms arise naturally in mathematics due to how they simplify the solution of exponential equations like the one in our exercise. When looking at the equation \( 250 e^{0.05 x}=400 \), taking the natural logarithm of both sides allows us to eliminate the exponential component, simplifying the process of solving for \( x \). This is possible because the property \( \ln(e^x) = x \) holds, meaning that the \( \ln \) of an \( e \) raised to any power will simply return that power.

Using natural logarithms in exponential equations allows us to handle problems where the variable is in the exponent. They're extremely beneficial for different fields such as calculus, because they make complex calculations more accessible.

Logarithmic Identities

Logarithmic identities are powerful tools that simplify the manipulation of logarithms. They provide shortcuts and simplify expressions where logarithms are involved. Two frequently used identities include the logarithm of an exponent and the change of base formula.

In our exercise, the identity \( \ln(e^x) = x \) is used. This lets us simplify the equation \( \ln(e^{0.05x}) \) to just \( 0.05x \). Such properties are incredibly useful when the exponent itself contains the variable we want to solve for, as it allows the variable to be extracted from the power.

Understanding these identities helps students solve logarithmic and exponential equations efficiently, as they often appear in math problems at various educational levels. These identities make solving logarithmic equations straightforward and manageable.

Solving Equations

Solving equations involves finding an unknown variable by manipulating the equation according to certain mathematical rules. When it comes to exponential equations, like in our scenario, you typically follow a systematic process:

• Isolate the exponential part to make the equation simpler.
• Apply the natural logarithm to both sides to eliminate the exponent.
• Use logarithmic identities to simplify the equation.
• Solve for the variable by isolating it on one side.

Let's apply this to our problem. First, isolate the term involving \( e^{0.05x} \) by dividing by 250. This gives \( e^{0.05x} = 1.6 \). Then use \( \ln \) to break down the exponent: \( 0.05x = \ln(1.6) \). Finally, solve for \( x \) by dividing by 0.05, simplifying the calculation enormously.

This systematic approach, applying natural logarithms and recognizing key logarithmic identities, is essential for solving exponential and logarithmic equations. With practice, you'll become proficient in using these strategies to tackle various types of equations.