Question:

To solve the given equation $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$, let's proceed step by step. Step 1: Observe the given equation and notice it resembles a quadratic equation if we let $e^x = t$ and $e^{2x} = t^2$. However, to make it clearer, we can try to factor it or find a suitable substitution that simplifies the equation. Step 2: Consider the equation as a quadratic in terms of $e^{2x}$. The equation can be rewritten as $(e^{2x})^2 + 8e^{2x} + 13 - 8e^x + 1 = 0$. However, this doesn't directly help. Instead, let's try to factor the original equation or find a pattern. Step 3: Looking closely, the equation can be seen as a quadratic in $e^{2x}$ if we group terms appropriately: $(e^{4x} + 13e^{2x} + 1) + 8e^{3x} - 8e^x = 0$. This doesn't immediately simplify. Let's consider a different approach by trying to factor or find a pattern that fits known factorable forms. Step 4: Recognize the equation can be seen as $(e^{2x} + 1)^2 + 8e^{2x}(e^x - 1) + 12 = 0$ after adding and subtracting appropriate terms to complete the square or factor. However, the direct path involves recognizing it as a quadratic in disguise by letting $z = e^{2x}$ and then trying to factor or apply a formula. Step 5: Let $z = e^{2x}$. The equation then transforms into $z^2 + 8z^{3/2} + 13z - 8z^{1/2} + 1 = 0$. This still doesn't simplify nicely. The correct approach involves recognizing the equation as a quadratic equation in a different form. Step 6: Correctly identify the equation as a quadratic in $e^{2x}$ by realizing the equation is symmetric and can be factored or solved by substitution. Let $t = e^x$, then the equation becomes $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$. This looks like a quadratic in $t^2$. Step 7: Now, consider the equation as $(t^2 + 1)^2 + 8t^2(t - 1) + 12 = 0$ or more directly as a quadratic by dividing through by $t^2$ and recognizing patterns. However, the equation $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$ can be approached by trying to factor it or find roots through rational root theorem or other means. Step 8: The equation can be factored or solved by recognizing it as $(t^2 - z_1)(t^2 - z_2) = 0$ where $z_1$ and $z_2$ are roots of a related quadratic equation. Let's correct the path and directly solve the given equation through an appropriate substitution. Step 9: Consider $z = e^x$ and rewrite the equation as $z^4 + 8z^3 + 13z^2 - 8z + 1 = 0$. This is a quartic equation. To simplify, notice it can be factored as $(z^2 - z - 1)(z^2 + z - 1) = 0$ by recognizing the pattern or using polynomial division. Step 10: Solve the quadratic equations $z^2 - z - 1 = 0$ and $z^2 + z - 1 = 0$. For $z^2 - z - 1 = 0$: $$ \begin{aligned} z &= \frac{1 \pm \sqrt{1 + 4}}{2} \\ z &= \frac{1 \pm \sqrt{5}}{2} \end{aligned} $$ For $z^2 + z - 1 = 0$: $$ \begin{aligned} z &= \frac{-1 \pm \sqrt{1 + 4}}{2} \\ z &= \frac{-1 \pm \sqrt{5}}{2} \end{aligned} $$ Step 11: Since $z = e^x$, the solutions for $z$ must be positive because $e^x > 0$ for all $x \in \mathbb{R}$. Thus, we only consider the positive roots: - From $z^2 - z - 1 = 0$, $z = \frac{1 + \sqrt{5}}{2}$. - From $z^2 + z - 1 = 0$, $z = \frac{-1 + \sqrt{5}}{2}$. Step 12: Now, solve for $x$ using $z = e^x$. For $z = \frac{1 + \sqrt{5}}{2}$ and $z = \frac{-1 + \sqrt{5}}{2}$, we have: $$ \begin{aligned} x &= \ln\left(\frac{1 + \sqrt{5}}{2}\right) \\ x &= \ln\left(\frac{-1 + \sqrt{5}}{2}\right) \end{aligned} $$ Both values of $z$ are less than $2$ (since $\sqrt{5} < 3$), implying $x = \ln(z) < \ln(2)$, and since $\ln(2) < 1$, both $x$ values are less than $1$ but more importantly, they are negative because both $z$ values are less than $1$. Therefore: 1 <div class="key-concept"><strong>Key Concept:</strong> Let $t = e^x > 0$. Divide the equation by $t^2$ to get $t^2 + 8t + 13 - \frac{8}{t} + \frac{1}{t^2} = 0$. Let $z = t - \frac{1}{t}$, reducing to $z^2 + 8z + 15 = 0$.</div> <div class="trap-box"><strong>Trap:</strong> Since $t = e^x > 0$, only positive values of $t$ are valid. Both resulting quadratics $t^2+3t-1=0$ and $t^2+5t-1=0$ have one positive root each, giving two real values of $x$, both negative since $0 < t < 1$.</div>