Gosh, i know the basic of exponentials, but when it contains some complex algebra stuff in there, i go blank.
I know to subtract the 10, but thats it...
$\displaystyle e^2x-7e^x + 10 = 0$ (its e to the 2x power)
Thanks a bunch!
Solve for x: $\displaystyle e^{2x}-7e^x + 10 = 0$ (you need to surround your exponent with braces if it has more than one item in it).
Let $\displaystyle u=e^x$, and then you have the quadratic $\displaystyle u^{2}-7u+ 10 = (u-2)(u-5)=0$, and therefore
u=5 or u=2
$\displaystyle e^x=5$ or $\displaystyle e^x=2$
Then x=ln 5 or x=ln 2
Do you mean "is there a harder way"? Of course, you don't have to replace $\displaystyle e^x$, that's just the simplest thing to do. You could recognise that $\displaystyle e^{2x}- 7e^x+ 10= (e^x- 2)(e^x- 5)= 0$ and then arguing that either $\displaystyle e^x- 2= 0$ or $\displaystyle e^x- 5= 0$.
And how do I recognise that factoring? Well, I recognize $\displaystyle e^{2x}- 7e^x+ 10$ as a "quadratic in $\displaystyle e^x$" which is essentially replacing $\displaystyle e^x$, at least in my head, by a single variable.