Hi I know what is the answer actually. for ln(x^2 - 1) = 3 ; x = \sqrt((1+e) (1-e+e^(2))), -\sqrt((1+e)(1-e+e^(2))) br/ And for e^2x - 3e^x + 2 = 0; x = ln(2), 0
But I don't really know how to get to these answers. I need to know the steps to get the answers.
Here are some suggestions.
For (a), you can exponentiate both sides; then you are solving a quadratic. (The exponential is a bijection, so you don't have to worry about introducing new solutions.)
For (b), if you write $y = e^x$ then you have $y^2 - 3y + 2 = 0$, which is another quadratic. Once you solve for $y$, you need to solve for $x$ via $x = \ln y$, being mindful of the domain of $\ln$. (It will turn out the domain doesn't matter for your particular question, but in principle it should.)