I have a function : $$f(x) = e^{3x}+4e^{2x}-3e^x=0$$ I have to show that there is exactly one real root.
My effort
$f$ is continuous on $\mathbb{R}$
$f(-1) = e^{-3}+4e^{-2}-3e^{-1} \approx =-0.512$
$f(0) = e^{0}+4e^{0}-3e^{0}=2$
So by the Intermediate Value Theorem, we know that there must be a number $c \in [−1, 0]$, where $f(c) = 0.$
So I call $c$ my solution to the original equation and suppose there exists another one; I call this one $d$. Then either $d > c$ or $d < c$. Then $f(x)$ is a continuous function on $[c,d]$ and differentiable on $(c,d)$ (it is differentiable everywhere, actually). By Rolle’s Theorem, there then must exist a point $r \in (c,d)$, where $f′(r) = 0$.
Doing so $$f'(x) = \mathrm{e}^x\cdot\left(3\mathrm{e}^{2x}+8\mathrm{e}^x-3\right)$$
substituting $u = e^x$ I take : $$ u(3u^2+8u-3)$$ that have two roots $u=1/3$ and $u=-3$ substituting back now I take $x= \log(1/3)$ and $x=-\log(3)$.I have two roots not one.
Am I correct ?