Problem 45 - College Algebra Essentials

Question

Solve each exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{4 x}+5 e^{2 x}-24=0$$

Step-by-Step Solution

Answer

The only solution is \( x \approx 0.55 \)

Step 1: Express as Quadratic Equation

Let \(y = e^{2x}\). Hence the equation \(e^{4 x}+5 e^{2 x}-24=0\) becomes \(y^2 + 5y - 24 = 0\)

Step 2: Use Quadratic Formula

Now, use the quadratic formula to solve for y, where \(a = 1\), \(b = 5\), \(c = -24\). Therefore, the solution for \(y\) will be \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - -96}}{2} = \frac{-5 \pm \sqrt{121}}{2} = \frac{-5 \pm 11}{2}\]

Step 3: Solve for y

Hence, the two solutions for \(y\) are \[y = 3, y = -8\]

Step 4: Back Substitution

Substitute \(y = e^{2x}\) back into the result: \(e^{2x} = 3\) and \(e^{2x} = -8\).

Step 5: Applying Natural Logarithms

We know that \(e^{2x} = -8\) has no solution because the exponential function is always positive. For the other solution: Apply the natural logarithm to both sides, \(ln(e^{2x}) = ln(3)\), which simplifies to \(2x = ln(3)\). Divide by 2: \(x = \frac{ln(3)}{2}\)

Step 6: Decimal Approximation

Finally, calculate a decimal approximation using a calculator: \( x \approx 0.55 \)