Question

The equation $e ^{4 x}+8 e ^{3 x}+13 e ^{2 x}-8 e ^x+1=0, x \in R$ has :

• A. four solutions two of which are negative
• B. two solutions and both are negative
• C. no solution
• D. two solutions and only one of them is negative

Hint:

When solving for exponential equations, ensure to recognize that the exponential function is always positive, which helps eliminate any non-real or invalid solutions.

Answer 1

The Correct Option is B

$e^{4x}+8e^{3x}+13e^{2x}-8e^x+1=0$
Let $e^x=t$
Now, $t^4+8t^3+13t^2-8t+1=0$
Dividing equation by $t^2$,
$t^2+8t+13-\frac{8}{t}+\frac{1}{t^2}=0$
$t^2+\frac{1}{t^2}+8\left(t-\frac{1}{t}\right)+13=0$
${\left(t-\frac{1}{t}\right)}^2+2+8\left(t-\frac{1}{t}\right)+13=0$
Let $t-\frac{1}{t}=z$
$z^2+8z+15=0$
$(z+3)(z+5)=0$
$z=-3$ or $z=-5$
So, $t-\frac{1}{t}=-3$ or $t-\frac{1}{t}=-5$
$t^2+3t-1=0$ or $t^2+5t-1=0$
$t=\frac{-3\pm \sqrt{13}}{2}$ or $t=\frac{-5\pm \sqrt{29}}{2}$
as $t=e^x$ so $t$ must be positive,
$t=\frac{\sqrt{13}-3}{2}$ or $\frac{\sqrt{29}-5}{2}$
So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$
Hence two solution and both are negative.
So , the correct option is (B) : two solutions and both are negative

Answer 2

Step 1: Let \( e^x = t \). Now the given equation becomes: \[ t^2 + 8t + 13t - 8t + 1 = 0. \] Step 2: Dividing the equation by \( t^2 \), we get: \[ t^2 + 8t + 13 - \frac{8}{t} = 0. \] Rewriting the equation: \[ \left( \frac{1}{t} \right)^2 + 2 \times 8 \times \left( \frac{1}{t} \right) + 13 = 0. \] Step 3: Let \( \frac{1}{t} = z \). Now the equation becomes: \[ z^2 + 8z + 15 = 0. \] Step 4: Solving the quadratic equation for \( z \): \[ z = -3 \quad \text{or} \quad z = -5. \] Step 5: Now solving for \( t \), we get: \[ t = -\frac{1}{3} \quad \text{or} \quad t = -\frac{1}{5}. \] Since \( t = e^x \), we know that \( e^x \) must be positive. Thus, we find: \[ x = \ln \left( \frac{1}{3} \right) \quad \text{or} \quad x = \ln \left( \frac{1}{5} \right). \] Step 6: Therefore, both solutions are negative. Hence, the correct answer is: \[ \boxed{ \text{Two solutions and both are negative.} } \]