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Math Help - 2logan = loga18 + loga ( n - 4 )

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    2logan = loga18 + loga ( n - 4 )

    Given that:

    2logan = loga18 + loga ( n - 4 )

    find the possible values of n.


    >>how do i get to logan^2 = loga 18(n-4)

    is it correct that

    2logan = 2 logan^2

    how do you get from:

    loga18 + loga ( n - 4 )

    to logan^2 = loga 18(n-4)
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  2. #2
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    Quote Originally Posted by ansonbound View Post
    Given that:

    2logan = loga18 + loga ( n - 4 )

    find the possible values of n.


    >>how do i get to logan^2 = loga 18(n-4)

    is it correct that

    2logan = 2 logan^2

    how do you get from:

    loga18 + loga ( n - 4 )

    to logan^2 = loga 18(n-4)
    I assume you mean base a.

    2\log_a(n) = \log_a(n^2) is true as it's the power law.

    You can simplify the RHS using the addition law:

    \log_a(18) + \log_a(n-4) = \log_a[18(n-4)]


    \log_a(n^2) = \log_a[18(n-4)]

    Now if the two bases are the same then the exponents must be equal.

    n^2 = 18(n-4)

    Which is a standard quadratic equation but remember that n>0 to satisfy the original domain
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