Key Concepts and Formulas

• Substitution: Replacing a complex expression with a single variable to simplify equations.
• AM-GM Inequality: For non-negative real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$.
• Quadratic Formula: The solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Divide by $e^{2x}$

We are given the equation $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$. We divide each term by $e^{2x}$ to exploit the symmetry of the coefficients. Since $e^{2x} > 0$ for all real $x$, this operation is valid and does not introduce extraneous solutions or remove any existing ones. $\frac{e^{4x}}{e^{2x}} + \frac{4e^{3x}}{e^{2x}} - \frac{58e^{2x}}{e^{2x}} + \frac{4e^x}{e^{2x}} + \frac{1}{e^{2x}} = 0$ Simplifying, we get: $e^{2x} + 4e^x - 58 + 4e^{-x} + e^{-2x} = 0$

Step 2: Group Terms and Substitute $t = e^x + e^{-x}$

We group the terms with reciprocal exponents: $(e^{2x} + e^{-2x}) + 4(e^x + e^{-x}) - 58 = 0$ Let $t = e^x + e^{-x}$. We want to express $e^{2x} + e^{-2x}$ in terms of $t$. We know that $(e^x + e^{-x})^2 = e^{2x} + 2e^x e^{-x} + e^{-2x} = e^{2x} + 2 + e^{-2x}$. Therefore, $e^{2x} + e^{-2x} = (e^x + e^{-x})^2 - 2 = t^2 - 2$. Substituting $t = e^x + e^{-x}$ and $e^{2x} + e^{-2x} = t^2 - 2$ into the equation, we have: $(t^2 - 2) + 4t - 58 = 0$ $t^2 + 4t - 60 = 0$

Step 3: Solve the Quadratic Equation for $t$

We solve the quadratic equation $t^2 + 4t - 60 = 0$. Factoring gives us $(t+10)(t-6) = 0$. Thus, $t = -10$ or $t = 6$.

Step 4: Determine the Valid Range of $t$

Since $t = e^x + e^{-x}$ and $e^x > 0$ for all real $x$, we can apply the AM-GM inequality to $e^x$ and $e^{-x}$: $\frac{e^x + e^{-x}}{2} \ge \sqrt{e^x \cdot e^{-x}}$ $\frac{t}{2} \ge \sqrt{1}$ $\frac{t}{2} \ge 1$ $t \ge 2$ Therefore, the valid range for $t$ is $[2, \infty)$. Since $t = -10$ is not in this range, we discard it. We are left with $t = 6$.

Step 5: Solve for $x$ using $t=6$

We have $e^x + e^{-x} = 6$. Let $p = e^x$. Then $e^{-x} = \frac{1}{p}$, and our equation becomes: $p + \frac{1}{p} = 6$ Multiplying by $p$, we get: $p^2 + 1 = 6p$ $p^2 - 6p + 1 = 0$ Using the quadratic formula: $p = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)}$ $p = \frac{6 \pm \sqrt{36 - 4}}{2}$ $p = \frac{6 \pm \sqrt{32}}{2}$ $p = \frac{6 \pm 4\sqrt{2}}{2}$ $p = 3 \pm 2\sqrt{2}$ So, $p_1 = 3 + 2\sqrt{2}$ and $p_2 = 3 - 2\sqrt{2}$. Since $3 = \sqrt{9}$ and $2\sqrt{2} = \sqrt{8}$, we have $3 > 2\sqrt{2}$, so both $p_1$ and $p_2$ are positive.

Step 6: Find the Number of Real Solutions for $x$

Since $p = e^x$, we have $x = \ln(p)$. Since we found two distinct positive values for $p$, we have two distinct real solutions for $x$: $x_1 = \ln(3 + 2\sqrt{2})$ $x_2 = \ln(3 - 2\sqrt{2})$

Common Mistakes & Tips

• Always check the range of the substituted variable after solving the simplified equation. Discard any solutions that fall outside the valid range.
• Remember the AM-GM inequality as a tool to determine the range of expressions like $e^x + e^{-x}$.
• Don't forget to check if the solutions for $p = e^x$ are positive, as the natural logarithm is only defined for positive numbers.

Summary

By dividing the original equation by $e^{2x}$ and using the substitution $t = e^x + e^{-x}$, we transformed the exponential equation into a quadratic equation. Solving for $t$ and considering the valid range $t \ge 2$, we found one valid value for $t$. Substituting this back and solving for $e^x$, we obtained two distinct positive values for $e^x$, which correspond to two distinct real solutions for $x$.

The final answer is \boxed{2}.