Question

If the sum of all the roots of the equation $e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$
is logeP, then p is equal to _____.

Answer 1

The correct answer is 45
Let $e^x = t$ then equation reduces to
$t^2−11t−\frac{45}{t}+\frac{81}{2}=0$
$⇒ 2t^3 – 22t^2 \+ 81t – 45 = 0 …(i)$
if roots of
$e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$
are α, β, γ then roots of (i) will be
$e^{α_1}e^{α_2}e^{α_3}$
Therefore , by using product of roots
$e^{α_1+α_2+α_3}=45$
$⇒ α_1 \+ α_2 \+ α_3$
= ln 45
⇒ p = 45