Solve for $x$ when $e^{2x}-3e^x=4$ 
Solve for $x$ when $e^{2x}-3e^x=4$

This is what I've gotten so far:
\begin{array}
he^{2x}-3e^x&=&4\\
\ln(e^{2x}-3e^x)&=&\ln(4) \\
\dfrac{\ln(e^{2x})}{\ln(3e^x)}&=&\ln(4)\\
\dfrac{2x}{\ln(3) + \ln(e^x)}&=&\ln(4) \\
\dfrac{2x}{\ln(3) + x}&=&\ln(4) \\
x= \dfrac{\ln(3) \cdot \ln(4)}{2-\ln(4)}
\end{array}
I clearly made a mistake somewhere, does anyone know where?
 A: $$\ln(e^{2x}-3e^x)\ne\dfrac{\ln e^{2x}}{\ln 3e^x}$$

Let $u=e^x$, this transforms the equation to a quadratic: $u^2-3u-4=0$ which can be used to obtain values for $u$ and eventually $x$.
A: Let $y=e^x$. Then $y^2-3y=4$. Solve this quadratic equation.
A: Substitute $u=e^x$ then you have to solve the quadratic $u^2-3u-4=0$
You also can do it this way:
$$e^{2x}-3e^x-4=(e^x+1)(e^x-4)=0\Longrightarrow e^x+1=0 \quad\text{or}\quad e^x-4=0$$.
A: Your mistake is that $\ln(3e^x)=\ln( 3) + x$, not $\ln(3)x$.
Another solution would be to make the substitution $y=e^x$ and solve the quadratic in $y$.
A: Write $$(e^x)^2-3e^x-4=0$$ and substitute $$t=e^x$$
A: Let, $e^x=y$ then the equation $e^{2x}-3e^x=4$ becomes $y^2-3y=4$
Solving $y^2-3y-4=0$ we get, $$\Delta=b^2-4ac=9+16=25$$ $$y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$\implies y=\frac{3\pm\sqrt{25}}{2}$$ $$\therefore y=-1,4$$
Substituting $y=e^x$
$e^x=-1\implies$ No Solution Exist $\forall x\in\mathbb R$ because $e^x>0$
So, $$e^x=4\implies x=\ln{4}$$
$$\therefore x=\ln{4}$$
