Problem 45 - Precalculus

Question

Solve for \(x:-350=211 e^{-0.025 x}-450\)

Step-by-Step Solution

Answer

The solution for \(x\) is approximately 30.152.

Step 1: Isolate the Exponential Term

The given equation is \(-350 = 211 e^{-0.025x} - 450\). To isolate the exponential term, add 450 to both sides of the equation, resulting in:\[-350 + 450 = 211 e^{-0.025x}\]This simplifies to:\[100 = 211 e^{-0.025x}\]

Step 2: Solve for the Exponential Expression

Divide both sides by 211 to solve for the exponential expression \(e^{-0.025x}\):\[\frac{100}{211} = e^{-0.025x}\]

Step 3: Take the Natural Logarithm

Take the natural logarithm of both sides of the equation to solve for \(-0.025x\):\[\ln\left(\frac{100}{211}\right) = \ln(e^{-0.025x})\]Recall that \(\ln(e^y) = y\), so:\[-0.025x = \ln\left(\frac{100}{211}\right)\]

Step 4: Solve for x

Divide both sides by \(-0.025\) to solve for \(x\):\[x = \frac{\ln\left(\frac{100}{211}\right)}{-0.025}\]

Step 5: Calculate the Numeric Result

Compute the values using a calculator. The natural logarithm value needed:\[\ln\left(\frac{100}{211}\right) \approx -0.7538\]Thus:\[x \approx \frac{-0.7538}{-0.025} = 30.152\]

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is an essential concept in mathematics, especially when dealing with exponential equations. It is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. When you take the natural logarithm of an exponential function, such as \( \ln(e^y) \), you directly obtain \( y \). This property makes natural logarithms incredibly useful for solving equations that contain the exponential function.

• If you have an equation involving \( e \), taking the natural logarithm on both sides will help you to "unwrap" the exponential function.
• In our case, \( \ln \left( e^{-0.025x} \right) = -0.025x \), using the property that \( \ln(e^y)=y \).
• This step effectively simplifies the problem, allowing us to isolate and solve for the variable \( x \).

Understanding natural logarithms not only helps in solving exponential equations but also provides insight into the rate of growth of processes, how different quantities scale, and much more in advanced mathematics and real-world applications.

Isolating Variables

Isolating variables is a fundamental algebraic technique where you rearrange an equation to express one variable in terms of others, often aiming to solve for that particular variable. This concept is crucial in solving any equation, exponential or otherwise. In the given exercise, our goal is to isolate \( x \). We start by isolating the exponential term \( e^{-0.025x} \), which involves:

• Adding or subtracting terms on both sides to move everything away from the variable of interest—initially, we add 450 to both sides.
• Dividing by any coefficients that are multiplied by the exponential term—here, dividing both sides by 211.

Once isolated, you can directly manipulate this term, such as using logarithms to simplify further, ultimately solving for the variable \( x \). Isolating the variable is often a crucial intermediate step that paves the way for simpler arithmetic manipulation and precise solutions.

Numerical Computation

In solving any mathematical problem, especially those involving exponential equations, numerical computation plays a key role. It involves using numbers and calculators for precise calculations after symbolic work is complete.After isolating the variable and transforming the equation into an easily solvable form, you often arrive at:

• Evaluating expressions, such as \( \ln \left( \frac{100}{211} \right) \) in this case, which approximately equals \( -0.7538 \).
• Performing arithmetic operations like division. For example, dividing \( -0.7538 \) by \( -0.025 \) gives roughly \( 30.152 \).

Numerical computation provides the final numeric result, ensuring that all components of the solution have been accurately calculated. This step, supported by technologies like calculators or computational software, ensures precision and helps to verify the validity of algebraic manipulations done earlier in the problem.