Thread: e to the y problem

Solving for y



Is this the way to go?



Make everything e to the something and then ln it all away.

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.

Originally Posted by Jason76

Solving for y



Is this the way to go?



Make everything e to the something and then ln it all away.

Then in it all away??? What does that even mean?

I suggest that you give us the original problem exactly and tell us what you are currently studying.

The OP wrote Ln (natural logarithm), not in...

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.

Agreed, because above there was a major algebraic mistake in trying to "make everything e to the something", and then "Ln it all" to normal digits that are easy to work with (ones with no "e" or "Ln" stuff.).

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.

Originally Posted by Prove It

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.

Were these three lines correct algebraically?

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.

??

Why are you beating a dead horse? IT CAN NOT BE DONE! Like I said, you need to get an approximate solution using a numerical method or technology.

Originally Posted by Prove It

Why are you beating a dead horse? IT CAN NOT BE DONE! Like I said, you need to get an approximate solution using a numerical method or technology.

Ok, sounds good. I'll try that. Anyhow, what is tip off that a problem needs a numerical method? How can a person spot such problems off the bat?

Generally when you see the variable you are trying to solve for both inside and outside of a transcendental function, then you can't solve for that variable analytically.