Problem 21 - College Algebra

Question

Solve each exponential equation in Exercises \(1-26\) Express the solution set in terms of natural logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{2 x}-3 e^{x}+2=0$$

Step-by-Step Solution

Answer

The solution to the equation \(e^{2x} - 3e^{x} + 2 = 0\) is \(x = ln(1)\) and \(x = ln(2)\), approximated as \(x \approx 0\) and \(x \approx 0.69\) respectively.

Step 1: Transform into a Quadratic Equation

The exponential equation \(e^{2x} - 3e^{x} + 2 = 0\) can be rewritten as a quadratic equation with the substitution \(e^{x} = y\). This yields a quadratic equation \(y^{2} - 3y + 2 = 0\).

Step 2: Solve the Quadratic Equation

The quadratic equation can be solved using the quadratic formula \(y = ( -b \pm \sqrt{b^2-4ac} ) / 2a\), where a, b, and c are coefficients from our equation \(y^{2} - 3y + 2 = 0\). So, \(a=1\), \(b=-3\), and \(c=2\). This results in solutions \(y_1 = 1\) and \(y_2 = 2\).

Step 3: Back Substitute

Since we made the substitution \(e^{x}=y\) we should now back substitute. The result yields two equations: \(e^{x}=1\) and \(e^{x}=2\).

Step 4: Solve for x

To find x for each equation, both sides of the equations can be taken the natural logarithm \(ln\). This will give \(x = ln(1)\) for the first equation and \(x = ln(2)\) for the second equation.

Step 5: Calculate Decimal Approximations

For the final step, the decimal approximation for each solution can be found using calculator. Thus we obtain \(x \approx 0\) for the first solution and \(x \approx 0.69\) for the second solution, correct to two decimal places.

Key Concepts

Natural Logarithms

Natural logarithms are an important concept in solving exponential equations. In algebra, when you have an equation involving the exponential function, such as \(e^x\), taking the natural logarithm usually helps isolate the variable \(x\). This is because natural logarithms (notated as \(\ln\)) are the inverse operation of exponentials with base \(e\).

• If you have \(e^x = a\), you can solve for \(x\) by taking the natural logarithm of both sides to get \(x = \ln(a)\).
• This step allows you to solve for \(x\) directly, which is especially helpful when dealing with complexities like those found in the transformation from quadratic equations involving exponential bases.

When solving problems such as the one provided, natural logarithms are used to deduce the value of \(x\) after back substituting the quadratic solutions into their exponential equivalents.

Quadratic Equation

The quadratic equation comes into play when transforming an exponential equation to a more familiar form. In the example given, the initial exponential equation \(e^{2x} - 3e^x + 2 = 0\) is transformed using substitution into a quadratic equation form. By letting \(e^x = y\), we restate the problem as \(y^2 - 3y + 2 = 0\). This makes it solvable using typical methods for quadratic equations.

• A quadratic equation is of the form \(ay^2 + by + c = 0\), where \(a\), \(b\), and \(c\) are constants.
• In this scenario, \(a=1\), \(b=-3\), and \(c=2\).

The solutions to the quadratic equation are found using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). By applying this formula, you find the roots that eventually help solve for \(x\) in the context of the original exponential equation.

Decimal Approximation

Decimal approximations are useful when exact solutions involve complex numbers or irrational numbers that cannot be easily expressed in a simpler form. They provide a practical way of presenting numerical solutions to a reasonable degree of accuracy, especially when using technology like calculators.In the exercise, after finding the logarithmic solutions \(x = \ln(1)\) and \(x = \ln(2)\), we aim for precision by converting these into decimal form:

• \(x = \ln(1)\) translates to \(x \approx 0.00\) because \(\ln(1) = 0\).
• \(x = \ln(2)\) results in \(x \approx 0.69\) when calculated to two decimal places.

This approach ensures that the final answers can be used in practical applications or further calculations where specific numerical values are required.

Substitution Method

The substitution method is a powerful tool that simplifies complex equations by temporarily replacing a difficult element with a simpler term. In the given exercise, substitution makes the exponential equation more manageable. The concept can be outlined as follows:

• You encounter a challenging expression: \(e^{2x} - 3e^x + 2 = 0\).
• We introduce a substitution \(e^x = y\), transforming it into \(y^2 - 3y + 2 = 0\), a straightforward quadratic equation.
• This substitution reduces complexity, allowing easy application of algebraic methods like the quadratic formula to solve for \(y\).
• After solving, replace the substitution back to its original form to find the solution in terms of the original variable.

Following this process aids in systematically breaking down and solving equations that initially appear daunting due to their format or expression.