Analyzing the Setup
The given equation is:
e4x+e3x−4e2x+ex+1=0
This is a quartic equation disguised in exponential form. To simplify the landscape, we let t=ex.
Since
ex is always positive, we must respect the domain constraint
t>0. The equation transforms into the following polynomial:
t4+t3−4t2+t+1=0
This is a reciprocal equation because the coefficients (1,1,−4,1,1) mirror each other perfectly.
The Reduction Strategy
To exploit this symmetry, we divide the entire equation by
t2. Since
t>0, we know
$t^2
eq 0$, making this operation valid:
t2+t−4+t1+t21=0
Next, we group the terms by their powers:
(t2+t21)+(t+t1)−4=0
The Bridge of u
Let
u=t+t1. Squaring this expression yields:
u2=t2+2+t21⇒t2+t21=u2−2
Substituting this into our grouped equation, the quartic expression reduces to a quadratic:
(u2−2)+u−4=0
u2+u−6=0
Factoring the quadratic gives:
(u+3)(u−2)=0
This provides two potential candidates:
u=−3 and
u=2.
Final Calculation and Verdict
We apply the
AM-GM inequality to the substitution
u=t+t1. For any
t>0:
The candidate
u=−3 fails this condition and is therefore rejected. We proceed with
u=2:
t+t1=2
t2−2t+1=0
(t−1)2=0
This yields t=1. Back-substituting ex=t, we get ex=1, which implies:
x=0
We have successfully navigated the fortress to find the single, elegant solution.