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Thread: Log question

  1. #1
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    Log question

    The problem:
    $\displaystyle
    \sqrt[7]{7}^{(4-3x)} = 7^{x^2}
    $


    Answer:
    $\displaystyle
    x = -1, \frac{4}{7}
    $

    How do you do this problem?

    It says the first thing I need to do is to get both sides to have the same base. Wouldn't the base be 10?
    $\displaystyle
    log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x
    $
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  2. #2
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    Quote Originally Posted by desiderius1 View Post
    The problem:
    $\displaystyle
    \sqrt[7]{7}^{(4-3x)} = 7^{x^2}
    $


    Answer:
    $\displaystyle
    x = -1, \frac{4}{7}
    $

    How do you do this problem?

    It says the first thing I need to do is to get both sides to have the same base. Wouldn't the base be 10?
    $\displaystyle
    log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x
    $
    base is $\displaystyle 7$ ...

    note that $\displaystyle \sqrt[7]{7} = 7^{\frac{1}{7}}$

    $\displaystyle \left(7^{\frac{1}{7}}\right)^{4-3x} = 7^{x^2}$

    bases equal ... exponents are equal

    $\displaystyle \frac{4-3x}{7} = x^2$

    solve for $\displaystyle x$
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  3. #3
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    $\displaystyle
    \frac{4-3x}{7} = x^2
    $

    $\displaystyle
    4-3x = 7x^2
    $

    $\displaystyle
    7x^2+3x-4=0
    $

    $\displaystyle
    (7x+1)(x-\frac{4}{7})
    $

    I get -4/7 and not -4, is this right?

    $\displaystyle
    7x^2-4x+x-\frac{4}{7}
    $

    $\displaystyle
    7x^2-3x-\frac{4}{7}
    $
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  4. #4
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    Quote Originally Posted by desiderius1 View Post
    $\displaystyle
    \frac{4-3x}{7} = x^2
    $

    $\displaystyle
    4-3x = 7x^2
    $

    $\displaystyle
    7x^2+3x-4=0
    $

    $\displaystyle
    (7x+1)(x-\frac{4}{7})
    $ ???
    $\displaystyle 7x^2 + 3x - 4 = 0$

    $\displaystyle (7x - 4)(x + 1) = 0$

    setting each factor equal to 0 ...

    $\displaystyle x = \frac{4}{7}$ , $\displaystyle x = -1$
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