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Math Help - Evaluating Logarithmics

  1. #1
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    Evaluating Logarithmics

    Solve

    27^{x} (9^{2x - 1}) = 3^{x + 4}

    My work:

    3^{3x} (3^{4x-2}) = 3^{x+4}

    (3x)(4x-2) = x+4

    12x^2 - 6x - x - 4 = 0

    12x^2 - 7x - 4 = 0

    Now I'm stuck, I can't find any factors that has the product of 48 and the sum of -7.


    Textbook Answer: 1
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by Macleef View Post
    Solve

    27^{x} (9^{2x - 1}) = 3^{x + 4}
    Divide by 3^x
    27/3 = 9, 3^{x + 4} = 3^x 3^4


    9^x(9^{2x - 1})=3^4

    Combine like bases of nine...

    9^{3x-1}=3^4

    Nine equals three squared...

    3^{6x-2}=3^4

    Now, take log based three of both sides...

    6x-2=4
    x=1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve

    27^{x} (9^{2x - 1}) = 3^{x + 4}

    My work:

    3^{3x} (3^{4x-2}) = 3^{x+4}

    \color{red}(3x)(4x-2) = x+4

    ...

    that is wrong. x^a \cdot x^b = x^{a {\color {red} +} b} NOT x^{ab}
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    Quote Originally Posted by Jhevon View Post
    that is wrong. x^a \cdot x^b = x^{a {\color {red} +} b} NOT x^{ab}
    I don't understand what you wrote...
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  5. #5
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    Quote Originally Posted by colby2152 View Post
    Divide by 3^x
    27/3 = 9, 3^{x + 4} = 3^x 3^4


    9^x(9^{2x - 1})=3^4

    Combine like bases of nine...

    9^{3x-1}=3^4

    Nine equals three squared...

    3^{6x-2}=3^4

    Now, take log based three of both sides...

    6x-2=4
    x=1
    Why is it common base of 9? I don't think you can get 27 from any exponents of 9?
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by Macleef View Post
    Why is it common base of 9? I don't think you can get 27 from any exponents of 9?
    9^x9^{(2x-1)} = 9^{(3x-1)}

    That is what I was referring to. As far as 27, that is equal to 3^3
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  7. #7
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    Quote Originally Posted by colby2152 View Post
    9^x9^{(2x-1)} = 9^{(3x-1)}

    That is what I was referring to. As far as 27, that is equal to 3^3
    What happens to the rest of 27? I still don't know how you get a common base of 9... (I know  3^{3} ) then wouldn't it be (3)9^{x} instead of just 9^{x}?
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    GAMMA Mathematics
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    Quote Originally Posted by Macleef View Post
    What happens to the rest of 27? I still don't know how you get a common base of 9... (I know  3^{3} ) then wouldn't it be (3)9^{x} instead of just 9^{x}?
    It starts out with 27^x. 27=3^3, so 27^x = 3^3x
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    Quote Originally Posted by colby2152 View Post
    Divide by 3^x
    27/3 = 9, 3^{x + 4} = 3^x 3^4


    9^x(9^{2x - 1})=3^4

    Combine like bases of nine...

    9^{3x-1}=3^4

    Nine equals three squared...

    3^{6x-2}=3^4

    Now, take log based three of both sides...

    6x-2=4
    x=1
    Okay, I finally get why it's 9^{x} ...

    Now, one more question, why is it just 3^{4}? What happened to to the "x" from 3^{x + 4}?
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  10. #10
    GAMMA Mathematics
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    Quote Originally Posted by Macleef View Post
    Okay, I finally get why it's 9^{x} ...

    Now, one more question, why is it just 3^{4}? What happened to to the "x" from 3^{x + 4}?
    3^{x+4}=3^x3^4
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  11. #11
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    Quote Originally Posted by colby2152 View Post
    3^{x+4}=3^x3^4
    what happened to 3^{x}?? I don't see it in your work...?? I don't see any common factoring if that's what you did...
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  12. #12
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    Quote Originally Posted by Macleef View Post
    what happened to 3^{x}?? I don't see it in your work...?? I don't see any common factoring if that's what you did...
    For any nonzero a:
    a^ma^n = a^{m + n}

    -Dan
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  13. #13
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    I still don't get it...but thanks for helping me anyways...
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    I still don't get it...but thanks for helping me anyways...
    Let's try an example:

    Let a = 3, m = 4, and n = 2. I'm going to calculate 3^4 \cdot 3^2.

    3^4 \cdot 3^2 = (3 \cdot 3 \cdot 3 \cdot 3) \cdot (3 \cdot 3) = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^6

    Our rule states that a^ma^n = a^{m + n} so
    3^4 \cdot 3^2 = 3^{4 + 2} = 3^6 (check!)

    Look at both ways of doing the problem and see if it helps you understand where this rule comes from. It's all about counting out the number of factors of "a" that you have.

    -Dan
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