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Math Help - Log question

  1. #1
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    Log question

    The problem:
    <br />
  \sqrt[7]{7}^{(4-3x)} = 7^{x^2}<br />


    Answer:
    <br />
x = -1, \frac{4}{7}<br />

    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?
    <br />
log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x<br />
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  2. #2
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    Quote Originally Posted by desiderius1 View Post
    The problem:
    <br />
  \sqrt[7]{7}^{(4-3x)} = 7^{x^2}<br />


    Answer:
    <br />
x = -1, \frac{4}{7}<br />

    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?
    <br />
log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x<br />
    base is 7 ...

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

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

    bases equal ... exponents are equal

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

    solve for x
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  3. #3
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    <br />
\frac{4-3x}{7} = x^2<br />

    <br />
4-3x = 7x^2<br />

    <br />
7x^2+3x-4=0<br />

    <br />
(7x+1)(x-\frac{4}{7})<br />

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

    <br />
7x^2-4x+x-\frac{4}{7}<br />

    <br />
7x^2-3x-\frac{4}{7}<br />
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  4. #4
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    Quote Originally Posted by desiderius1 View Post
    <br />
\frac{4-3x}{7} = x^2<br />

    <br />
4-3x = 7x^2<br />

    <br />
7x^2+3x-4=0<br />

    <br />
(7x+1)(x-\frac{4}{7})  <br />
???
    7x^2 + 3x - 4 = 0

    (7x - 4)(x + 1) = 0

    setting each factor equal to 0 ...

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