Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-9 e^{x}-36=0\)

Short answer

The approximate solution to the equation \(e^{2x} - 9e^x - 36 = 0\) to three decimal places is \(x = 2.485\).

Step-by-step solution

Define the Substitution

Since the equation looks like a quadratic, we will substitute \(e^x\) with the variable \(y\). So, the equation becomes \(y^2 - 9y - 36 = 0\)

Solve the Quadratic Equation

The quadratic equation can be solved using the quadratic formula: \(y = [ -(-9) \pm \sqrt{(-9)^2 - 4 * 1 * -36} ] / 2*1 \). So, \( y = [9 \pm \sqrt{81+144}] / 2 = [9 \pm \sqrt{225}] / 2 = [9 \pm 15] / 2 \). Therefore, the solutions are \(y= -3\) and \(y= 12\)

Substitute Back and Solve for x

Now, substitute back the original variable which was defined as \(y=e^x\) . We obtain \(e^x = -3\) and \(e^x = 12\). The left side of the first equation, \(e^x = -3\), cannot be negative for real number \(x\), hence this equation has no solutions. The solution to the second equation, \(e^x = 12\), is, by taking the natural logarithm on both sides, \(x = \ln(12)\). The approximate value to three decimal places is \(x = 2.485\)

Key concepts

Quadratic Substitution

When dealing with complex exponential equations, sometimes employing a technique known as **quadratic substitution** can simplify the process. This method is particularly useful when the equation resembles a quadratic form. For example, in the equation \(e^{2x} - 9e^{x} - 36 = 0\), by recognizing that \(e^{2x} = (e^{x})^2\), we can use the substitution \(y = e^{x}\). This transforms the exponential equation into a standard quadratic equation: \(y^2 - 9y - 36 = 0\).
This conversion is a powerful move because it allows us to apply familiar techniques from quadratic equations, such as the quadratic formula, to find solutions.

• This method reorganizes the equation so that it takes a more manageable form.
• It allows for the use of quadratic formulas or factoring techniques.
• However, remember to substitute back the original variable for the final solution.

Natural Logarithm

The **natural logarithm**, denoted as \(\ln\), is a vital tool when working with exponential equations, especially when solving for variables in expressions like \(e^x = 12\).
This function is the inverse of the exponential function \(e^x\), which means it "undoes" the effect of an exponential expression. In essence, \(\ln(e^x) = x\).

So, in our converted quadratic equation after solving for \(y = e^x = 12\), the natural logarithm helps us to isolate \(x\) by taking \(\ln\) on both sides: \(x = \ln(12)\).

• The natural logarithm is particularly handy for converting exponential expressions into linear ones.
• It converts difficult-to-solve exponential problems into simpler algebraic forms.
• It can only take positive arguments, as the logarithm of a negative number is undefined in the real numbers.

Approximation Methods

Once you have the expression \(x = \ln(12)\), sometimes finding an exact numerical answer isn't feasible or necessary, and using **approximation methods** becomes essential. This is where technology or approximating techniques step in.
Approximating \(\ln(12)\) involves calculating its value numerically to reach the precision required by the problem, in this case, three decimal places.

• Creating an approximation can be achieved using calculators or computational software.
• Approximations are important when values can't be expressed in simple rational forms.
• In real-world applications, knowing how to round appropriately is crucial to get the nearest practical value.

In practice, this would round \(x = \ln(12)\) to approximately \(x = 2.485\), capturing an accurate enough picture for most analyses.