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  1. #1
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    Post solve algebraically

    2log(base4)X - log(base4)(x+3)=1
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by joeyisnice123 View Post
    2log(base4)X - log(base4)(x+3)=1
    Note that 2\log_4x-\log_4\left(x+3\right)=1\implies \log_4\left(x^2\right)-\log_4\left(x+3\right)=1

    Now recall that \log_ab-\log_ac=\log_a\left(\frac bc\right)

    Thus, \log_4\left(x^2\right)-\log_4\left(x+3\right)=1\implies\log_4\left(\frac{  x^2}{x+3}\right)=1

    Now, note that \log_44=1

    Thus, we now have the equation \log_4\left(\frac{x^2}{x+3}\right)=\log_44

    Since we want the two logarithms to be equal, we must see where \frac{x^2}{x+3}=4

    Can you continue on from here?

    --Chris
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