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Math Help - Exponents and Logarithms

  1. #1
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    Exponents and Logarithms

    1.Solve, 3^2x = 5^(1-x)


    2. Solve 4^2x - 8(2^2x) - 9 =0
    Last edited by nav92; October 30th 2010 at 11:27 PM. Reason: Sorry i mistyped on question 1..it's 3^2x not 5^2x..
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  2. #2
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    Take logarithms to the base 5 on both sides. You have 2x = 1-x which is a simple linear equation in one variable.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    For the second one, use the substitution:

    y = 2^{2x}
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  4. #4
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    For the second one,

    The answer must be in 3 significant numbers which I can't seem to get......Instead I get,

    After substitution,

    y^2 - 8y - 9 = 0

    y= 9 ; y= -1
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Right, now, replace y by 2^2x again
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  6. #6
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    Oh man , how could i missed the substitution part on question 2...okay i get it
    What about question 1 where the base number is different?
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Hm... let's do it.

    3^{2x} = 5^{1-x}

    Take logs on both sides.

    log3^{2x} = log5^{1-x}

    2x log3 = (1-x)log5

    2x log3 = log5 - xlog5

    Bring xlog5 to the left;

    2x log3 + x log 5 = log5

    Factor x;

    x (2log3 +  log 5) = log5

    Can you continue now?
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  8. #8
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    Quote Originally Posted by nav92 View Post
    Oh man , how could i missed the substitution part on question 2...okay i get it
    What about question 1 where the base number is different?
    Re-write the equation:

    3^{2x} = 5^{1-x}~\implies~9^x=\dfrac5{5^x}

    Multiply through by 5^x and you'll get:

    9^x \cdot 5^x = 5~\implies~(9 \cdot 5)^x=5

    I'll leave the rest for you.
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  9. #9
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    Yes this had really helped a lot.....Question solved.
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