Problem 63 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, solve for \(x\) or \(t\). $$ 4 e^{2 x-1}-1=5 $$

Step-by-Step Solution

Answer

Therefore the solution for x is \(x = \frac{\ln(1.5) + 1}{2}\)

Step 1: Isolate the Exponential Term

First, you need to isolate the exponential term on one side of the equation. You can do this by adding 1 to both sides of the equation. \[4 e^{2 x-1}=5+1\] \[4 e^{2 x-1}=6\]

Step 2: Express the Exponent as a Single Term

Next, divide each side by 4 to isolate \(e^{2 x-1}\) on one side of the equation. This gives us: \[e^{2 x-1} = \frac{6}{4} = 1.5\]

Step 3: Take Natural Logarithm (ln) of both sides

Since the base of the exponential is e (natural logarithm base), you can take the natural logarithm of both sides. When you do this, the exponent comes out front due to properties of logarithms. So, we get: \[2x-1 = \ln(1.5)\]

Step 4: Solve for x

Lastly, to isolate \(x\), add 1 to both sides of the equation and then divide by 2. This provides the solution: \[x = \frac{\ln(1.5) + 1}{2}\]

Key Concepts

Isolating Exponential Terms

To solve exponential equations like the one in our exercise, the first crucial step is to isolate the exponential term. Begin by focusing on the exponential term on one side of the equation. In our problem, the equation is \(4 e^{2x-1} - 1 = 5\). Here, the exponential term involves \(e^{2x-1}\), which is multiplied by 4 and then reduced by 1.

To start isolating, add 1 to both sides of the equation to eliminate the subtraction operation, resulting in:

• \(4 e^{2x-1} = 6\)

Next, divide both sides by 4 to undo the multiplication, leaving you with the exponential alone:

• \(e^{2x-1} = 1.5\)

This sets the stage for applying further algebraic operations. By successfully isolating \(e^{2x-1}\), you've made the problem much simpler to work with in subsequent steps.

Natural Logarithm

Once the exponential term \(e^{2x-1}\) is isolated, the next logical move involves the natural logarithm (ln). Logarithms, particularly the natural kind due to their base \(e\), are terrific at dealing with exponential equations because they allow the removal of the exponent. This property is pivotal in our exercise.

Take the natural logarithm of both sides of the equation \(e^{2x-1} = 1.5\). When you do so, you leverage the logarithmic property that enables the exponent to "come down":

• \(\ln(e^{2x-1}) = \ln(1.5)\)

Due to the property of logarithms where \(\ln(e^y) = y\), your equation simplifies to:

• \(2x - 1 = \ln(1.5)\)

Thus, applying the natural logarithm has transformed an exponential equation into a linear form that is ready to be solved.

Solving for Variables

With the equation now transformed to \(2x - 1 = \ln(1.5)\), the final task is to solve for the variable \(x\). This step involves simple algebraic manipulations and is often the most straightforward.

The aim is to isolate \(x\), which involves undoing the operations affecting it. Start by adding 1 to both sides of the equation to offset the "-1":

• \(2x = \ln(1.5) + 1\)

Once this is completed, the equation is now \(2x = \ln(1.5) + 1\). Now divide each side by 2 to isolate \(x\):

• \(x = \frac{\ln(1.5) + 1}{2}\)

And there you have it! You've found the value of \(x\) by applying simple algebraic techniques. This solution leverages understanding how to rearrange equations and utilize logarithmic properties effectively.