I am super confused how does this step end up with this?
Then this is the working, I dont understand the second step please help me to show the missing step any Law of Logarithms at work here?
I am super confused how does this step end up with this?
Then this is the working, I dont understand the second step please help me to show the missing step any Law of Logarithms at work here?
In the second step, both sides of the equation are multiplied by $e^{3x}$. In the third step, $2 \cdot e^{3x}$ is subtracted from both sides of the equation. It can be rephrased as a quadratic equation in $y$ where $y = e^{-3x}$ and $y \in \mathbb{R^{+}}$.
Perhaps this might be easier to follow using substitution.
With the substitution, $y=e^{3x}$ (forcing $y>0$), this becomes $ \begin{array}{rrll} &y-8/y&=2&(\text{substitution})\\ \therefore&y^2-2y-8&=0&(\text{algebra})\\ \therefore&y&=1\pm3&(\href{http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula}{\text{quadratic formula}}) \end{array} $ Therefore, $e^{3x}=4\Rightarrow x=\frac23\log(2)$.