Results 1 to 12 of 12
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - e to the y problem

  1. #1
    Super Member
    Joined
    Oct 2012
    From
    USA
    Posts
    699
    Thanks
    11

    e to the y problem

    Solving for y

    y^{2} - 9 = e^{y}

    Is this the way to go?

    (e^{y})(y^{2}) - (e^{y})(9) = e^{y}(e^{y})

    Make everything e to the something and then ln it all away.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576

    Re: e to the y problem

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    770
    Thanks
    391

    Re: e to the y problem

    Quote Originally Posted by Jason76 View Post
    Solving for y

    y^{2} - 9 = e^{y}

    Is this the way to go?

    (e^{y})(y^{2}) - (e^{y})(9) = e^{y}(e^{y})

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576

    Re: e to the y problem

    The OP wrote Ln (natural logarithm), not in...
    Thanks from JeffM
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2012
    From
    USA
    Posts
    699
    Thanks
    11

    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.

    y^{2} - 9 = e^{y}

    (e^{y})(y^{2}) - (e^{y})(9)} = (e^{y})(e^{y})

    ye^{y + 2}} - 9e^{y} = e^{2y}

    \ln[ye^{y + 2} - 9e^{y}] = \ln[e^{2y}}]

    y + 2 - 9y = 2y

    -8y + 2 = 2y

    2 = 10y

    y = \dfrac{1}{5} ??

    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.).
    Last edited by Jason76; June 14th 2014 at 10:04 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Oct 2012
    From
    USA
    Posts
    699
    Thanks
    11

    Re: e to the y problem

    Quote Originally Posted by Prove It View Post
    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?

    ye^{y + 2}} - 9e^{y} = e^{2y}

    \ln[ye^{y + 2} - 9e^{y}] = \ln[e^{2y}}]

    y + 2 - 9y = 2y
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576

    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Oct 2012
    From
    USA
    Posts
    699
    Thanks
    11

    Re: e to the y problem

    y^{2} - 9 = e^{y}

    (e^{y})(y^{2}) - (e^{y})(9)} = (e^{y})(e^{y})

    e^{y}y^{2} - 9e^{y} = e^{2y}

    \ln[e^{y}y^{2} - 9e^{y}] = \ln[ e^{2y}]

    \ln[\dfrac{e^{y}y^{2}}{9e^{y}}] = \ln[ e^{2y}]

    ??
    Last edited by Jason76; June 14th 2014 at 10:29 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576

    Re: e to the y problem

    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Oct 2012
    From
    USA
    Posts
    699
    Thanks
    11

    Re: e to the y problem

    Quote Originally Posted by Prove It View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,802
    Thanks
    1576

    Re: e to the y problem

    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.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum