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Math Help - Equations with the exponential function.

  1. #1
    Member Furyan's Avatar
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    Equations with the exponential function.

    Hello

    I know this is elementary, but I don't know how solve this:

    The question is, find the exact value(s) of x which satisfy the equation.

    e^{2x} = e^x +12

    I am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.

    I know if the question was:

    e^{2x} = 12

    Then the answer would be:

     x = \dfrac{1}{2}\ln12

    Some help would be very much appreciated. Thank you.
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  2. #2
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    Re: Equations with the exponential function.

    Quote Originally Posted by Furyan View Post
    Hello

    I know this is elementary, but I don't know how solve this:

    The question is, find the exact value(s) of x which satisfy the equation.

    e^{2x} = e^x +12

    I am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.

    I know if the question was:

    e^{2x} = 12

    Then the answer would be:

     x = \dfrac{1}{2}\ln12

    Some help would be very much appreciated. Thank you.
    e^{2x} - e^x - 12 = 0

    (e^x - 4)(e^x + 3) = 0

    finish it
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  3. #3
    Member Furyan's Avatar
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    Re: Equations with the exponential function.

    Dear Skeeter

    Quote Originally Posted by skeeter View Post
    e^{2x} - e^x - 12 = 0

    (e^x - 4)(e^x + 3) = 0

    finish it
    A quadratic in e, that was left field and I wasn't looking for it. Should have seen it though.

     x = \ln4 = 2\ln2

    e^x = -3 has no solutions.

    Thank you very much indeed.
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  4. #4
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    Re: Equations with the exponential function.

    Hopefully, you know that e^{2x}= (e^x)^2. If you let y= e^x, the equation becomes y^2= y+ 12 which is the same as y^2- y- 12= 0, a quadratic equation in y. If you did not notice that this could be factored as (y- 4)(y+ 3), you could solve it by completing the square or using the quadratic formula.
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  5. #5
    Member Furyan's Avatar
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    Re: Equations with the exponential function.

    Hello HallsofIvy

    Quote Originally Posted by HallsofIvy View Post
    Hopefully, you know that e^{2x}= (e^x)^2. If you let y= e^x, the equation becomes y^2= y+ 12 which is the same as y^2- y- 12= 0, a quadratic equation in y. If you did not notice that this could be factored as (y- 4)(y+ 3), you could solve it by completing the square or using the quadratic formula.
    Thank you. I do find letting y= e^x and solving the quadratic in y easier to get my head round. I was able to factor that one.
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  6. #6
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    Re: Equations with the exponential function.

    Hello, Furyan!

    Find the exact value(s) of x which satisfy the equation."

    . . e^{2x} - e^x -12\:=\:0

    As skeeter pointed out, this is a quadratic equation.

    There is a way to recognize this phenomenon.

    There are usually three terms, written in standard order.

    If the first term has twice the exponent of the second term,
    . . we may have a quadratic.

    In your problem: . e^{2(x)} - e^x - 12 \:=\:0


    Another example: . x^6 - 7x^3 - 8 \:=\:0

    Note that we have: . \left(x^3\right)^2 - 7(x^3) - 8 \:=\:0

    Let u \,=\,x^3

    Then we have: . u^2 - 7u - 8 \:=\:0
    . . (u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1

    Back-substitute: . \begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}

    And there are four complex roots as well.

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  7. #7
    Member Furyan's Avatar
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    Re: Equations with the exponential function.

    Hello Soroban

    Quote Originally Posted by Soroban View Post
    Hello, Furyan!


    As skeeter pointed out, this is a quadratic equation.

    There is a way to recognize this phenomenon.

    There are usually three terms, written in standard order.

    If the first term has twice the exponent of the second term,
    . . we may have a quadratic.

    In your problem: . e^{2(x)} - e^x - 12 \:=\:0


    Another example: . x^6 - 7x^3 - 8 \:=\:0

    Note that we have: . \left(x^3\right)^2 - 7(x^3) - 8 \:=\:0

    Let u \,=\,x^3

    Then we have: . u^2 - 7u - 8 \:=\:0
    . . (u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1

    Back-substitute: . \begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}

    And there are four complex roots as well.

    Thank you for that. I will be sure to look out for this sort of equation in the future.
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