Problem 17 - Technical Mathematics with Calculus

Question

Solve for \(x\) to three significant digits. $$7 e^{1.5 x}=2 e^{2.4 x}$$

Step-by-Step Solution

Answer

Using a calculator, \(x \approx \frac{\ln(3.5)}{0.9}\) which computes to approximately 0.999.

Step 1: Write down the given equation

Start with the equation provided: \(7 e^{1.5x} = 2 e^{2.4x}\).

Step 2: Isolate the exponential terms

Divide both sides of the equation by \(2 e^{1.5x}\) to isolate the variables on one side. The equation becomes \(\frac{7}{2} = e^{(2.4x - 1.5x)}\) which simplifies to \(\frac{7}{2} = e^{0.9x}\).

Step 3: Take the natural logarithm of both sides

To eliminate the exponential, take the natural logarithm (ln) of both sides. This gives us \(\ln(\frac{7}{2}) = \ln(e^{0.9x})\). Applying the property \(\ln(e^y) = y\), we can simplify the right side to \(0.9x\).

Step 4: Solve for x

Now we have \(\ln(\frac{7}{2}) = 0.9x\). Divide by 0.9 to isolate x: \(x = \frac{\ln(\frac{7}{2})}{0.9}\).

Step 5: Calculate the value of x

Using a calculator, find the value of \(x\) to three significant digits: \(x \approx \frac{\ln(3.5)}{0.9}\).

Key Concepts

Natural Logarithm

The natural logarithm, commonly denoted as \textbf{ln}, is a mathematical function that is the inverse of the natural exponential function. For a positive number \(a\), the natural logarithm of \(a\) is the power to which the base \(e\) (Euler's number, approximately equal to 2.71828) must be raised to obtain \(a\). In simpler terms, if we have an equation \(e^x = a\), then \(x = \textbf{ln}(a)\).

The natural logarithm is particularly useful in solving exponential equations because it helps to 'undo' the effect of the \(e\) (exponential) function, thus enabling the isolation of the variable in question. For instance, the property \(\textbf{ln}(e^y) = y\) is crucial in transforming an equation from its exponential form to a linear one, making it easier to solve. This method was showcased in the step-by-step solution of the given exercise where the natural logarithm was applied to both sides of the equation to simplify and solve for the variable \(x\).

What's really valuable is understanding that the natural logarithm isn't just a random operation but is profoundly connected to rates of growth and decay, compounding processes, and much more, which are common in various scientific and financial fields.

Isolating Variables

When solving for a variable, one common goal is to \textbf{isolate} that variable on one side of the equation. Isolating the variable means manipulating the equation in such a way that the variable stands alone on one side, and everything else is on the opposite side. To do this, you generally perform operations that 'undo' the operations affecting the variable. These can include addition, subtraction, multiplication, division, and other operations specific to certain functions, such as taking logarithms for exponentials.

In the context of the given exercise, we were dealing with an exponential equation. The variable \(x\) was wrapped up in an exponent, making it difficult to isolate with basic algebraic operations. Dividing both sides of the equation by one of the exponential terms aided in this process, simplifying the equation to a point where the natural logarithm could be utilized to further isolate \(x\). This clever manipulation is an essential skill, as isolating the variable is often the most critical step towards solving any algebraic equation.

Exponential Functions

An \textbf{exponential function} is a mathematical expression in which a constant base is raised to a variable exponent. The general form of an exponential function is \(f(x) = b^x\), where \(b\) is the base and \(x\) is the exponent. The most common base for an exponential function in higher mathematics is \(e\), Euler's number, because of its unique properties in calculus, especially concerning growth and decay processes.

Exponential functions are characterized by rapid growth or decay. They appear frequently in real-world phenomena such as population growth, radioactive decay, and interest calculation in finance. In the context of the exercise, the terms \(e^{1.5x}\) and \(e^{2.4x}\) represent exponential functions with the base \(e\). Solving equations with exponential functions often involves utilizing logarithms—natural logarithms in the case of base \(e\)—to transform the equation into a solvable linear form. This transformation allows us to remove the variable from the exponent and facilitates the isolation of the variable, thereby enabling us to solve the equation.