1) ln(1/sqrt(1+x^2)) +1=0
???
2) e^x=1+2e^-x
for this one I've got e^(x)-2e=1, then e(x-2)=1. I'm just unsure of where to go from here?
1) $\displaystyle \ln\left(\frac{1}{\sqrt{1+x^2}} \right) + 1 = 0$
$\displaystyle \ln\left(\frac{1}{\sqrt{1+x^2}} \right) = -1$
change to an exponential equation ...
$\displaystyle e^{-1} = \frac{1}{\sqrt{1+x^2}}$
solve for $\displaystyle x$
2) $\displaystyle e^x = 1 + 2e^{-x}$
multiply every term by $\displaystyle e^x$ ...
$\displaystyle e^{2x} = e^x + 2$
$\displaystyle e^{2x} - e^x - 2 = 0$
factor the quadratic and solve for $\displaystyle x$
well, now you've seen it ...
I wanted to get rid of the $\displaystyle e^{-x}$ in the equation.
it's like if you had the equation $\displaystyle y = 1 + \frac{2}{y}$ ... you'd multiply every term by $\displaystyle y$ to clear the unwanted fraction.
... and it makes the equation quadratic in $\displaystyle e^x$, which can be easily factored and solved.