Animated Solution for Mathematics - Differentiation: The number of real roots of the equation $e^{4x}+2e^{3x}-e^x-6=0$ is:

The equation $e^{4x}+2e^{3x}-e^x-6=0$ appears complex due to the exponential terms. However, recognizing the structure of $e^{4x}$, $e^{3x}$, and $e^x$ allows us to treat this as a hidden polynomial.

We define the substitution $t=e^x$. Since the range of the exponential function is strictly positive, we must enforce the boundary condition $t>0$.

Substituting $t$ into the original equation yields the fourth-degree polynomial:

To analyze the roots of this polynomial, we examine its behavior using calculus. The first derivative, which represents the slope of the function, is given by:

To determine the monotonicity of $f(t)$, we look at the second derivative:

For all $t>0$, $f^{\prime \prime }(t)$ is strictly positive. This implies that $f^{\prime }(t)$ is a strictly increasing function for $t>0$.

Because $f^{\prime }(0)=-1$ and $f^{\prime }(1)=9$, the Intermediate Value Theorem guarantees that $f^{\prime }(t)$ crosses zero exactly once at some value $\alpha \in (0,1)$. This point $\alpha$ represents the global minimum of the function $f(t)$ for $t>0$.

We now apply the Intermediate Value Theorem to the function $f(t)$ itself to locate the root:

Since $f(t)$ is continuous and changes sign between $t=1$ and $t=2$, there exists exactly one root in the interval $(1,2)$.

Because $t=e^x$ and we have established there is exactly one positive root $t$, we conclude that there is exactly one real solution for $x$. You have successfully stripped away the exponential layer to reveal the underlying polynomial structure.