You won't be able to get exact solutions for this equation. You will have to use numerical method of solution, like the Bisection Method or Newton's Method.

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- June 14th 2014, 04:55 PM #1

- June 14th 2014, 05:38 PM #2

- June 14th 2014, 06:43 PM #3

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- June 14th 2014, 06:45 PM #4

- June 14th 2014, 08:37 PM #5
## Re: e to the y problem

The solution should be -3 because that's where the two functions intersect. Maybe it was a number a little less or more than -3, but it was somewhere near there. Actually it can't be -3 cause plugging it into the equations does not make them equal.

??

Sorry if some big mistake here. Probably there is.

You won't be able to get exact solutions for this equation. You will have to use numerical method of solution, like the Bisection Method or Newton's Method.

- June 14th 2014, 08:57 PM #6
## Re: e to the y problem

There is basically nothing about that attempted solution which is correct. Like I said, there is no point trying to solve it analytically when the variable to be solved for is both inside and outside of a function, like it is here.

y = -3 is clearly not a solution, since the LHS is $\displaystyle \begin{align*} (-3)^2 - 9 = 0 \end{align*}$ and the RHS is $\displaystyle \begin{align*} \mathrm{e}^{-3} \neq 0 \end{align*}$. These are clearly not the same thing.

- June 14th 2014, 09:02 PM #7

- June 14th 2014, 09:11 PM #8
## Re: e to the y problem

First of all, $\displaystyle \begin{align*} \mathrm{e}^y\,y^2 \neq y\,\mathrm{e}^{y + 2} \end{align*}$, so the first line is invalid.

Taking logarithms of both sides is ok, but $\displaystyle \begin{align*} \ln{ \left( y\,\mathrm{e}^{y + 2} - 9\mathrm{e}^y \right) } \neq y + 2 - 9y \end{align*}$. Go back and review your logarithm laws. $\displaystyle \begin{align*} \log{(a+b)} \neq \log{(a)} + \log{(b)} , m\log{(n)} \neq \log{(mn)} \neq \log{(n)} \end{align*}$. You have essentially tried to simplify using these three fallacies.

- June 14th 2014, 09:25 PM #9

- June 14th 2014, 09:27 PM #10

- June 14th 2014, 09:34 PM #11

- June 14th 2014, 09:42 PM #12