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Math Help - Logarithm help need explanation

  1. #1
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    Logarithm help need explanation

    Write 1 - 2log7x as a single logarithm.
    Thanks!
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  2. #2
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    Re: Logarithm help need explanation

    Quote Originally Posted by dzomberg View Post
    Write 1 - 2log7x as a single logarithm.
    Thanks!

    \log_7(7)-\log_7(x^2)=\log_7\left(\frac{7}{x^2}\right)~.
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  3. #3
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    Re: Logarithm help need explanation

    $$1{\rm{ }} - {\rm{ }}2{\log _7}x = {\log _7}7 - {\log _7}{x^2} = {\log _7}{7 \over {{x^2}}}$$
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  4. #4
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    Re: Logarithm help need explanation

    Hi, this kind of problems involves the use of properties. So you want that expression to be simplified in one logarithm.

    First, let's write everything as a logarithm. We know that: log_n (n) = 1 so we could say that  n = 7 (in our case). Thus  1 = log_7 (7)

    Then, we have to put, for the second part, "inside" the logarithm. So we also know that: a \cdot log_n(x) = log_n(x^a). Finally 2 \cdot log_7(x) = log_7(x^2)

    We put it altogether (using the substraction of logarithm property of the same base) log_n(x)-log_n(y) = log_n\left(\frac{x}{y}\right):

     1 - 2\cdot log_7(x) = log_7(7) - log_7(x^2) = log_7\left(\frac{7}{x^2}\right)
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