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

Question

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?

Answer 1

$$\ln(e^{2x}-3e^x)\ne\dfrac{\ln e^{2x}}{\ln 3e^x}$$

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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 $x$.

Answer 2

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 $y$.

Answer 3

Let $y=e^x$. Then $y^2-3y=4$. Solve this quadratic equation.

Answer 4

Substitute $u=e^x$ then you have to solve the quadratic $u^2-3u-4=0$

You also can do it this way:

$$e^{2x}-3e^x-4=(e^x+1)(e^x-4)=0\Longrightarrow e^x+1=0 \quad\text{or}\quad e^x-4=0$$.

Answer 5

Write $$(e^x)^2-3e^x-4=0$$ and substitute $$t=e^x$$

Answer 6

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, $$e^x=4\implies x=\ln{4}$$

$$\therefore x=\ln{4}$$