Question

The number of real solutions of the equation e4x + 4e3x - 58e2x + 4ex + 1 = 0 is _____.

Answer 1

The given equation e4x + 4e3x - 58e2x + 4ex + 1 = 0
Dividing by e2x
e2x + 4ex – 58 + 4e–x + e–2x = 0
⇒ (ex + e–x)2 + 4(ex + e–x) – 60 = 0
Let ex + e–x = t ∈ [2, ∞)
⇒ t2 + 4t – 60 = 0
⇒ t = 6 is only possible solution
ex + e–x = 6
⇒ e2x – 6ex + 1 = 0
Let ex = p,
p2 – 6p + 1 = 0
$⇒ p = \frac {3+\sqrt 5}{2}$ or, $\frac {3-\sqrt 5}{2}$
Therefore,
$x = ln (\frac {3+\sqrt 5}{2})$

or, $x=ln (\frac {3-\sqrt 5}{2} )$

So, The number of real solutions of the equation is 2.