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    • Thread Starter
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    3^t+1 = 6 + 3^2t-1

    I just guessed on a calculator and it was 1.
    The answer is, how?
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    Just to clarify, is your equation:

     3^{t+1} = 6 + 3^{2t-1} ?

    If it is then note:

     (a^{m})^{n} = a^{mn},

    as well as:  (a^{m})(a^{n}) = a^{m+n} and  (a^{m})(a^{-n}) = a^{m-n}.

    Can you let let:  y = 3^{t},

    then re-write your original equation to reveal a familiar type of equation?
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    (Original post by simon0)
    Just to clarify, is your equation:

     3^{t+1} = 6 + 3^{2t-1} ?
    Yes
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    Nvm. Much more complicated method which I said and wont work always.

    do what poster above said.
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    Just to clarify, I had re-written my post above to include a method.
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    (Original post by y.u.mad.bro?)
    Start off by taking log of both side. What do you get?
    ln [3^(t+1)] - ln [3^(2t-1)] = 6
    ln [3^(t+1) / 3^(2t-1)] =6
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    (Original post by Sakura-Sama)
    ln [3^(t+1)] - ln [3^(2t-1)] = 6
    ln [3^(t+1) / 3^(2t-1)] =6
    Sorry, I thought it wa easier to use logs but it actually isn't. You are better off realising that it is a disguised quadratic.
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    (Original post by Sakura-Sama)
    ln [3^(t+1)] - ln [3^(2t-1)] = 6
    ln [3^(t+1) / 3^(2t-1)] =6
    Note, generally:

     \log(a+b) \neq \log(a) + \log(b), where  a, b \in \mathbb{R} .
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    I tried working this out as well but I can't seem to get to the solution?
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    it is a disguised quadratic... but you need an extra step than in typical examples. you need to split the 3{2t -1} so the 3{2t } appears without the -1
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    (Original post by user20167)
    I tried working this out as well but I can't seem to get to the solution?

    What if I stated:

     3^{t+1} = 3(3^{t}), and  3^{2t-1} = (3^{-1})(3^{2t}).

    Does this help?
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    I got the quadratic equation 1/3y^2 + 6 -3y=0
    is this right?
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    (Original post by user20167)
    I got the quadratic equation 1/3y^2 + 6 -3y=0
    is this right?
    Looks good, can you finish it off?
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    (Original post by simon0)
    Looks good, can you finish it off?
    I get the solutions 3 and 6 which don't work when put back into the equation though
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    (Original post by user20167)
    I get the solutions 3 and 6 which don't work when put back into the equation though
    Remember your substitution. What was "y"?
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    (Original post by simon0)
    Remember your substitution. What was "y"?
    1.63 and 1 are the solutions
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    (Original post by user20167)
    1.63 and 1 are the solutions
    Looks good.
 
 
 
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