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  1. #1
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    Logs

    SOLVE FOR X:

     <br />
\log_{8}{2^3}+\log_{3}\left(x^2-8\right) = \log_{4}\sqrt{16}+\log_{3}\left(2\right)+2\log_{9}  {x}<br />
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Sakai View Post
    SOLVE FOR X:

     <br />
\log_{8}{2^3}+\log_{3}\left(x^2-8\right) = \log_{4}\sqrt{16}+\log_{3}\left(2\right)+2\log_{9}  {x}<br />
    Hint: \log_a a=1 and \log_{a^2}x=\tfrac{1}{2}\log_a x. Do some minor simplifications first and try to get everything in terms of log base 3.

    Can you try to proceed?
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  3. #3
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    Cool! I didn't know that [LaTeX ERROR: Convert failed] .
    Quote Originally Posted by Chris L T521 View Post
    Hint: \log_a a=1 and \log_{a^2}x=\tfrac{1}{2}\log_a x. Do some minor simplifications first and try to get everything in terms of log base 3.
    Can you try to proceed?
    So \log_{8}{2^3} = 1, \log_{4}\sqrt{16} = 1, and 2\log_{9}{x} = 2\log_{3^2}{x} = \frac{2}{2}\log_{3}{x} = \log_{3}{x}. I get \log_{3}\left(x^2-8\right) = \log_{3}{x}+\log_{3}{2} \Leftrightarrow \log_{3}\left(x^2-8\right) = \log_{3}{2x} which means 2x = x^2-8, so we solve the quadratic x^2-2x-8 = (x-4)(x+2) = 0 to get x = 4 and x = -2. Thank you.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Sakai View Post
    Cool! I didn't know that [LaTeX ERROR: Convert failed] .
    Its a consequence of the change of base formula: \log_{a^2}x=\frac{\ln x}{\ln(a^2)}=\frac{\ln x}{2\ln a}=\frac{1}{2}\left(\frac{\ln x}{\ln a}\right)=\tfrac{1}{2}\log_a x

    So \log_{8}{2^3} = 1, \log_{4}\sqrt{16} = 1, and 2\log_{9}{x} = 2\log_{3^2}{x} = \frac{2}{2}\log_{3}{x} = \log_{3}{x}. I get \log_{3}\left(x^2-8\right) = \log_{3}{x}+\log_{3}{2} \Leftrightarrow \log_{3}\left(x^2-8\right) = \log_{3}{2x} which means 2x = x^2-8, so we solve the quadratic x^2-2x-8 = (x-4)(x+2) = 0 to get x = 4 and x = -2. Thank you.
    But note that x=-2 can't be a solution! Do you see why?
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    Its a consequence of the change of base formula: \log_{a^2}x=\frac{\ln x}{\ln(a^2)}=\frac{\ln x}{2\ln a}=\frac{1}{2}\left(\frac{\ln x}{\ln a}\right)=\tfrac{1}{2}\log_a x<br />
    Thanks. I will need to familiarise myself with that formula.
    But note that x=-2 can't be a solution! Do you see why?
    Because \log{\left(x\right)} is defined only for positive x, right?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Sakai View Post
    Because \log{\left(x\right)} is defined only for positive x, right?
    Correct!
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