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Math Help - Exponential equation

  1. #1
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    Exponential equation

    Hi,

    I am really stuck on how you deal with with an exponential equation something like this:

    x^y - 5 * x^y = 0 I know I cant multiply 5 * x^y as they have a different base (right?)

    How do i simplify this and get the variables/exponents together?

    Thanks
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  2. #2
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    Re: Exponential equation

    Quote Originally Posted by andy000 View Post
    Hi,

    I am really stuck on how you deal with with an exponential equation something like this:

    x^y - 5 * x^y = 0 I know I cant multiply 5 * x^y as they have a different base (right?)

    How do i simplify this and get the variables/exponents together?

    Thanks
    Good morning,

    is this the correct equation?

    x^y - 5x^y = 0

    If so, collect like terms:

    -4x^y = 0~\implies~x^y = 0

    There is only one real solution : x = 0~ \wedge~ y \ne 0
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  3. #3
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    Re: Exponential equation

    That is not the correct equation.

    The actual equation i am working on is:

    16^x - 5 * 8^x = 0

    I have tried converting into base of 2 so:

    (2^4)^x - 5 * (2^3)^x = 0

    = 2^4x - 5 * 2^3x = 0

    but I have no idea how to deal with the -5 * part of the equation.

    Thanks
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  4. #4
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    Re: Exponential equation

    Quote Originally Posted by andy000 View Post
    That is not the correct equation.

    The actual equation i am working on is:

    16^x - 5 * 8^x = 0

    I have tried converting into base of 2 so:

    (2^4)^x - 5 * (2^3)^x = 0

    = 2^4x - 5 * 2^3x = 0

    but I have no idea how to deal with the -5 * part of the equation.

    Thanks
    Hello,

    you can factor the term on the LHS:

    16^x - 5 \cdot 8^x = 0 ~\implies~2^{3x} \left(2^x-5 \right)=0

    A product is zero if one factor is zero. Since 2^{3x} \ne 0 only

    2^x-5 = 0~\implies~2^x=5~\implies~x = \log_2(5)
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  5. #5
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    Re: Exponential equation

    Hi, thanks very much!

    When you write 2^3x does not equal 0 do you mean 2^3x could not equal 0 for any value of x thus we can't get a real solution from that part of the equation?

    Thanks again
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  6. #6
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    Re: Exponential equation

    another approach
    16^x= 5*8^x
    xlog 16 =log 5 + xlog8
    using base 2
    4x=log 5 +3x
    x=log base 2 (5)
    x = ln5 /ln2
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  7. #7
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    Re: Exponential equation

    Quote Originally Posted by andy000 View Post
    Hi, thanks very much!

    When you write 2^3x does not equal 0 do you mean 2^3x could not equal 0 for any value of x thus we can't get a real solution from that part of the equation?

    Thanks again
    Yes.

    $x\ is\ a\ real\ number \implies 2^x > 0 \implies \left(2^x\right)^3 > 0 \implies 2^{3x} > 0 \implies 2^{3x} \ne 0.$

    I think the easiest way to deal with this is:

    $16^x - 5 * 8^x = 0 \implies 16^x = 5 * 8^x \implies$

    $\left(2^4\right)^x = 5 * \left(2^3\right)^x \implies$

    $2^x = 5 \implies$

    $ln(2^x) = ln(5) \implies$

    $x = \dfrac{ln(5)}{ln(2)}.$

    Edit: The answer in terms of log base 2 is more succinct, but it does not readily lead to a numeric approximation whereas an answer in terms of log base e can easily be computed on a scientific calculator.
    Last edited by JeffM; March 17th 2014 at 11:00 AM.
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