The number of real roots of the equation $$e^{4x} - e^{3x} - 4e^{2x} - e^x + 1 = 0$$ is equal to _________

We start with the given transcendental equation

$$e^{4x}-e^{3x}-4e^{2x}-e^{x}+1=0.$$

Because every power of $$e^{x}$$ appears, it is natural to put

$$t=e^{x}\qquad\text{so that}\qquad t>0\;(\text{since }e^{x}>0\text{ for every real }x).$$

Substituting $$t$$ for $$e^{x}$$ in each term we obtain a pure polynomial equation:

$$t^{4}-t^{3}-4t^{2}-t+1=0.$$

Denote this polynomial by $$g(t)=t^{4}-t^{3}-4t^{2}-t+1.$$ We must find how many positive real roots $$g(t)=0$$ possesses, because each positive root $$t_{0}$$ will correspond to a real root $$x_{0}=\ln t_{0}$$ of the original equation.

Step 1 - Maximum possible number of positive roots.
We recall Descartes’ Rule of Signs: the number of positive real roots of a polynomial with real coefficients does not exceed the number of sign-changes in the sequence of its coefficients, and differs from it by an even integer.

The coefficients of $$g(t)$$ are

$$+1,\;-1,\;-4,\;-1,\;+1,$$

giving the sign pattern $$+,-,-,-,+.$$ We see
  • the first change $$+\to -$$ (one change),
  • then $$-\to -$$ (no change),
  • again $$-\to -$$ (no change),
  • finally $$-\to +$$ (a second change).

Thus there are exactly two sign-changes; therefore the polynomial can have at most $$2$$ or $$0$$ positive real roots.

Step 2 - Locating sign changes of the polynomial.
We evaluate $$g(t)$$ at several convenient positive points.

• At $$t\to0^{+}$$ the dominant term is the constant $$+1$$, so

$$\lim_{t\to0^{+}}g(t)=+1\;>\;0.$$

• At $$t=1$$ we get

$$g(1)=1-1-4-1+1=-4\;<\;0.$$

Because the value changes from positive near $$t=0$$ to negative at $$t=1$$, the Intermediate Value Theorem guarantees at least one root in the interval $$(0,1).$$

• Next we try $$t=2$$:

$$g(2)=16-8-16-2+1=-9\;<\;0.$$

• Then $$t=3$$:

$$g(3)=81-27-36-3+1=16\;>\;0.$$

The change from negative at $$t=2$$ to positive at $$t=3$$ gives another sign-crossing, so there is at least one root in the interval $$(2,3).$$

Step 3 - Concluding the count.
We have located

• one positive root in $$(0,1),$$ and
• one positive root in $$(2,3).$$

Thus at least two positive roots exist. But, by Descartes’ Rule, there cannot be more than two. Therefore there are exactly two positive real roots of the polynomial equation $$g(t)=0$$, and hence exactly two real roots of the original exponential equation.

So, the answer is $$2$$.