Question

The number of real roots of the equation \( e^{4x} + 2e^{3x} - e^x - 6 = 0 \) is :

• A. 0
• B. 1
• C. 2
• D. 4

Hint:

For equations involving \( e^x \), transform to a polynomial in \( t \) and focus only on \( t>0 \). Negative roots of the polynomial do not provide real solutions for \( x \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We substitute \( t = e^x \) to convert the exponential equation into a polynomial equation. Since \( e^x \) is always positive for any real \( x \), we only look for positive real roots of the resulting polynomial.
Step 2: Detailed Explanation:
Let \( t = e^x \), where \( t>0 \). The equation becomes:
\[ P(t) = t^4 + 2t^3 - t - 6 = 0 \]
Applying Descartes' Rule of Signs to \( P(t) \):
- The signs of the coefficients are \( (+1, +2, -1, -6) \).
- There is exactly one sign change (from \( +2 \) to \( -1 \)).
- According to the rule, there is exactly one positive real root for \( t \).
Let's check the function behavior:
- At \( t = 1 \), \( P(1) = 1 + 2 - 1 - 6 = -4<0 \).
- At \( t = 2 \), \( P(2) = 16 + 16 - 2 - 6 = 24>0 \).
By the Intermediate Value Theorem, a root exists between \( 1 \) and \( 2 \). Since there is only one positive root for \( t \), there is only one real value \( x = \ln t \) that satisfies the original equation.
Step 3: Final Answer:
The number of real roots is 1.