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Thread: Logarithm

  1. #1
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    Logarithm

    Hi everyone... need some logarithms questions.... Thanks a lot for your help in advance. I appreciate it greatly.

    Question 1
    Prove that $\displaystyle \log_{y} x = \frac{1}{log_{x} y}$
    Given that x and y are real positive numbers.

    Find the possible values of x if $\displaystyle 2(\log_ {9} x + \log_{x} 9) = 5 $

    Question 2
    If $\displaystyle a^2 + b^2 = 7ab$
    show that $\displaystyle 2\log_{10} \frac{a + b}{3} = \log_{10} a + log_{10} b $
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  2. #2
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    Quote Originally Posted by ecogreen View Post
    Hi everyone... need some logarithms questions.... Thanks a lot for your help in advance. I appreciate it greatly.

    Question 1
    Prove that $\displaystyle \log_{y} x = \frac{1}{log_{x} y}$
    Given that x and y are real positive numbers.

    Find the possible values of x if $\displaystyle 2(\log_ {9} x + \log_{x} 9) = 5 $

    Question 2
    If $\displaystyle a^2 + b^2 = 7ab$
    show that $\displaystyle 2\log_{10} \frac{a + b}{3} = \log_{10} a + log_{10} b $

    1. Using the change of base rule...

    $\displaystyle \log_y{x} = \frac{\ln{x}}{\ln{y}}$

    $\displaystyle = \frac{1}{\frac{\ln{y}}{\ln{x}}}$

    $\displaystyle =\log_x{y}$.


    Using this information:

    $\displaystyle 2(\log_ {9} x + \log_{x} 9) = 5$

    $\displaystyle \log_9{x} + \log_x{9} = \frac{5}{2}$

    $\displaystyle \log_9{x} + \frac{1}{\log_9{x}} = \frac{5}{2}$

    $\displaystyle X + \frac{1}{X} = \frac{5}{2}$, where $\displaystyle X = \log_9{x}$

    $\displaystyle X^2 + 1 = \frac{5}{2}X$

    $\displaystyle X^2 - \frac{5}{2}X + 1 = 0$

    $\displaystyle 2X^2 - 5X + 2 = 0$

    $\displaystyle 2X^2 - 4X - X + 2 = 0$

    $\displaystyle 2X(X - 2) - 1(X - 2) = 0$

    $\displaystyle (X - 2)(2X - 1) = 0$

    $\displaystyle X - 2 = 0$ or $\displaystyle 2X - 1 = 0$

    $\displaystyle X = -2$ or $\displaystyle X = \frac{1}{2}$.


    So that means $\displaystyle \log_9{x} = -2$ or $\displaystyle \log_9{x} = \frac{1}{2}$.

    $\displaystyle x = 9^{-2}$ or $\displaystyle x = 9^{\frac{1}{2}}$

    $\displaystyle x = \frac{1}{81}$ or $\displaystyle x = 3$.
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