Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(1000 e^{-4 x}=75\)

Step-by-Step Solution

Answer

The solution to the equation is approximately \(x=0.723\).

Step 1: Isolate the term with the variable exponent

First, we should isolate the term with the variable exponent. We can do this by dividing both sides of the equation by 1000:$\frac{1000 e^{-4 x}}{1000}=\frac{75}{1000}$ which simplifies to $e^{-4 x}=0.075$.

Step 2: Apply natural logarithm to both sides

Next, we take the natural logarithm of both sides of the equation: $\ln(e^{-4 x})=\ln(0.075)$. Because the natural logarithm and 'e' are inverse functions, we simplify to $-4x=\ln(0.075)$.

Step 3: Solve for the variable

Finally, we solve the equation for the variable 'x'. We do this by dividing both sides of the equation by -4: $x= \frac{\ln(0.075)}{-4}$.

Step 4: Approximate the result

By entering $\frac{\ln(0.075)}{-4}$ into a calculator, the result is approximately $x=0.723$ to three decimal places.

Key Concepts

Natural Logarithm

The natural logarithm, often represented as \(\ln\), is a logarithm that uses the base \(e\), where \(e\) is an irrational number approximately equal to 2.71828. In mathematics, the natural logarithm is particularly useful for solving exponential equations because it acts as the inverse of the exponential function involving \(e\).

• Logarithms answer the question: "To what power must the base be raised, to obtain a specific number?"
• In the expression \(\ln(e^{-4x})\), the natural logarithm simplifies because \(\ln(e^x) = x\).

Using the property that \(\ln(e^a) = a\), we can convert more complex exponential equations into simpler, linear ones. This is especially helpful when dealing with equations where the exponent is a variable.

Variable Exponent

A variable exponent means that the variable, often denoted as \(x\), is used as the exponent in an exponential function. For the equation \(1000 e^{-4x} = 75\), the \(-4x\) is the variable exponent.

• Exponential functions with variable exponents can model diverse phenomena such as population growth or radioactive decay.
• Unlike linear equations, solving exponential equations with variable exponents often involves logarithms to "bring down" the exponent.
• Covering the steps: isolating the exponent, then using logarithms, can be a great technique to solve the problem effectively.

In our exercise, by taking the logarithm of both sides of the isolated equation, we were able to reframe it from the form \(e^{-4x} = 0.075\) to a linear equation involving \(\ln(0.075)\), making it easier to solve for \(x\).

Algebraic Solution

To solve an exponential equation algebraically means you use algebraic techniques and properties to find the unknown variable. This process often includes several clear, logical steps.
Here's a simplified breakdown of what we did in the solution:

• **Isolate the term containing the variable exponent**: We divided each side by 1000 to make this happen first.
• **Apply a logarithm to both sides**: Using the natural logarithm, we took advantage of its inverse ability with the base \(e\).
• **Solve for the variable**: By simplifying, we found \(\ln(0.075)\) and divided by \(-4\) to isolate \(x\).

This algebraic solution is precise and allows for the approximation of the result using a calculator. That's how we arrived at \(x \approx 0.723\), clearly showing how exponential equations can be untangled using algebra.