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Math Help - Solve the equation

  1. #1
    Senior Member Paze's Avatar
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    Solve the equation

    Hi MHF.

    I am asked to solve the equation 3^{2x-1}+27=10\cdot3^x

    I can pretend that 3^x=t and solve for t to get t1=27\\t2=3 however the book states that x1=1\\x2=3 I don't know how to solve for x.. Only t..So far !

    Perhaps MHF can clear this one up for me! Also, am I in the right section with this problem, or is this simply algebra? Thank you!
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  2. #2
    MHF Contributor

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    Re: Solve the equation

    You were the one who decided to make t= 3^x! And once you found that t= 27 and t= 1 are roots, you should immediately realize that 3^x= 27 and 3^x= 3. Isn't it clear what x must be?

    (If it were something more complicated, say 3^x= 25, you could use a logarithm: log(3^x)= xlog(3)= log(25) and then x= \frac{log(25)}{log(3)} where the logarithm can be to any convenient base. But since both "3" and "27" are powers of 3, you were probably not intended to do that.)
    Thanks from Paze
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  3. #3
    Senior Member Paze's Avatar
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    Re: Solve the equation

    Quote Originally Posted by HallsofIvy View Post
    You were the one who decided to make t= 3^x! And once you found that t= 27 and t= 1 are roots, you should immediately realize that 3^x= 27 and 3^x= 3. Isn't it clear what x must be?

    (If it were something more complicated, say 3^x= 25, you could use a logarithm: log(3^x)= xlog(3)= log(25) and then x= \frac{log(25)}{log(3)} where the logarithm can be to any convenient base. But since both "3" and "27" are powers of 3, you were probably not intended to do that.)
    Oh my. I didn't realize that. The answer was right in front of me! I guess I am slowly moving into the logarithm method you posted so thanks for that too!
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  4. #4
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    Re: Solve the equation

    Hello, Paze!

    \text{Solve: }\:3^{2x-1}+27\:=\:10\cdot3^x

    Multiply by 3: . 3^{2x} + 81 \:=\:30\cdot3^x

    Let t = 3^x
    Then we have: . t^2 + 81 \:=\:30t \quad\Rightarrow\quad t^2 - 30t + 81 \:=\:0

    . . (t-3)(t-27) \:=\:0 \quad\Rightarrow\quad \begin{Bmatrix}t \,=\,3\\t\,=\,27\end{Bmatrix}


    Back-substitute: . \begin{Bmatrix}3^x \:=\:3 & \Rightarrow& x \:=\:1 \\ 3^x \:=\:27 & \Rightarrow & x \:=\:3 \end{Bmatrix}

    Thanks from topsquark
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