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Math Help - Need help!! test in 7 hours!!!

  1. #1
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    Need help!! test in 7 hours!!!

    solve for x

    8(5^2x) + 8(5^x) = 6


    ALSO!

    (3^5x)(9^x^2)= 27 solve for x
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  2. #2
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    #1:
    \begin{aligned} 6 & = 8 \cdot 5^{2x} + 8 \cdot 5^x \\ 0 & = 8 \left({\color{red}5^x}\right)^2 + 8 \cdot {\color{red}5^x} - 6 \\ 0 & = 4\left({\color{red}5^x}\right)^2 + 4\left({\color{red}5^x}\right) - 3\end{aligned}

    This is basically a quadratic. To see it more easily, let u = 5^x.

    Then our quadratic becomes: 0 = 4u^2 + 4u - 3

    Factor to solve for u .

    __________

    #2: Is this your question: \left(3^{5x}\right)\left(9^{x^2}\right) = 27

    If so, try to get everything under the same base using your exponent laws:
    \begin{aligned} \left(3^{5x}\right)\left( (3^2)^{x^2}\right) & = 27 \\ \left(3^{5x}\right)\left(3^{2x^2}\right) & = 3^3 \\ 3^{2x^2 + 5x} & = 3^3 \end{aligned}

    Since the bases are equal, then the exponents should be equal, i.e. 2x^2 + 5x = 3
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  3. #3
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    Indices and logs

    Hi -

    Quote Originally Posted by capitalh View Post
    solve for x

    8(5^2x) + 8(5^x) = 6
    5^{2x} = 5^{x+x}=5^x \times 5^x=(5^x)^2

    So the equation can be re-writtenL

    8(5^x)^2+8(5^x)-6=0

    Substitute u=5^x to get a quadratic in u. Solve and then use \text{ln }u=x \text{ln }5 to find x.


    (3^5x)(9^x^2)= 27 solve for x
    Re-write everything as a power of 3:

    3^{5x}.(3^2)^{x^2}=3^3

    \implies 3^{5x}.3^{2x^2}=3^3

    ... can you do the next step? [Hint: 3^a.3^b=3^{a+b}]

    You then get a quadratic in x which you can solve.

    Grandad
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  4. #4
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    thanks guys!
    Last edited by mr fantastic; December 18th 2008 at 01:49 AM. Reason: Deleted reference to a useless (and hence deleted) reply
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