(note: that's e to the negative x power, i just don't know how to show that, sorry.)
How do you solve for x here?
BTW: If you have more than one character in an exponent, set off the whole exponent in braces.
[tex]e^{-x}[/tex] gives $\displaystyle e^{-x} $ instead of $\displaystyle e^-x $.
$\displaystyle e^x+e^{-x}=3$ multiply by $\displaystyle e^x$.
$\displaystyle e^{2x}-3e^x+1=$.
Can you solve $\displaystyle w^2-3w+1=0?$
If so let $\displaystyle z=e^x$.
Originally Posted by DarthBlood
errr....no sorry.
May I, with all due respect, ask this question?
If you cannot do a basic beginning algebra question, then why are you being asked to do an advanced algebra question involving exponentials?