Solve for X: e^x(6-e^X)=9
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Originally Posted by VNVeteran Solve for X: e^x(6-e^X)=9 Distributing and rearranging, we get $\displaystyle e^{2x}-6e^x+9=0$. Do you notice anything nice about this expression (don't think too hard about it)?
From where Chris left off consider a substitution $\displaystyle a = e^x$ giving $\displaystyle a^{2}-6a+9=0$ Now you only need to solve a qudartic
Originally Posted by pickslides From where Chris left off consider a substitution $\displaystyle a = e^x$ giving $\displaystyle a^{2}-6a+9=0$ Now you only need to solve a qudartic (a-3) (a-3) a=3 a=3 now what?
Originally Posted by VNVeteran (a-3) (a-3) a=3 a=3 now what? $\displaystyle a=e^x$, so $\displaystyle e^x=3\implies x=\ldots$
it factors to $\displaystyle (e^x - 3)(e^x - 3)$ $\displaystyle e^x = 3$ take $\displaystyle ln$ to both sides $\displaystyle lne^x = ln3$ $\displaystyle lne$ cancels out so you have $\displaystyle x = ln3$
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