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Math Help - Substituting equation

  1. #1
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    Substituting equation

    I'm having trouble substituting this equation ....

    {e}^{2t} - {243}^{t} + 143 = 0

    substitute x = {e}^{t}

    Giving answer to 4 decimal places

    Any help would be appreciated ...
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  2. #2
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    are you sure that you typed that correctly?

    {e}^{2t} - {243}^{t} + 143 = 0

    is hard

    {e}^{2t} - {243}e^{t} + 143 = 0

    can be solved using the subsititution you suggested.

    x^2 - 243x + 143 = 0

    ... (quadratic formula)...

    x = 243.587 (you will need more decimals when you do it)
    or
    x = -0.58 (ignore this negative solution. Can you see why?)


    Finally:
    since x = e^t

    243.57 = e^t

    ln(243.57) = t
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  3. #3
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    {e}^{2t} - {243}^{t} + 143 = 0

    Are you sure the problem is correctly written?
    The problem may be
    {e}^{2t} - {243}*{e}^{t} + 143 = 0
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  4. #4
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    For any real number, a, a= e^{ln(a)} so, in particular, 243^t= e^{ln(243^t}}= e^{t ln(243)}= (e^t)^{243} so if the equation you give is corect, then you can write it as e^{2t}+ 243^t+ 143= (e^t)^2+ (e^t)^{ln(243)}+ 143= x^2+ x^{ln(243)}+ 143.
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  5. #5
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    So sorry everyone fo this ....

    the actual equation was supposed to be

    {e}^{2t} - {24e}^{t} + 143 = 0 ...... subing x = {e}^{t}

    So very sorry
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  6. #6
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    From the laws of exponents you know that: e^{2t} = (e^t)^2

    Thus when you make your sub you get x^2-24x+143=0.

    Solve using your favourite method (it will factor but it's probably quicker to use the formula)
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  7. #7
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    Quote Originally Posted by daemonlies View Post
    So sorry everyone fo this ....

    the actual equation was supposed to be

    {e}^{2t} - {24e}^{t} + 143 = 0 ...... subing x = {e}^{t}

    So very sorry
    I think e^(*pi) got it right, but is that middle term (24e)^t or 24e^t?

    -Dan
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  8. #8
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    hi dan ... thanks for this ... its supposed to be .... 24e^t
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  9. #9
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    Assuming that it's e^(2t) - 24e^t + 143 = 0, make the substitution x = e^t, giving

    x^2 - 24x + 143 = 0.

    Now solve the quadratic and then use that solution to solve for t.
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