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Thread: Logarithm Problems

  1. #1
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    Logarithm Problems

    Can someone help me?

    1. $\displaystyle \log_{4}x - \log_{16}(x+3) = \frac{1}{2}$
    2. $\displaystyle \log(x+12) + \log(x-3) = 2$
    3. $\displaystyle \log_{x^2}9 - \log_{x}36 = 1 $


    And for this problem, can you check if I solved it right:
    $\displaystyle \log_{x}2 + \log_{x}3 = 5 $
    $\displaystyle \log_{x}(2 * 3) = 5$
    $\displaystyle \log_{x}6 = 5$
    $\displaystyle 6 = x^5$
    $\displaystyle 1.431 = x$



    Thank you
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  2. #2
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    Quote Originally Posted by meiyukichan View Post
    Can someone help me?
    [*]$\displaystyle \log_{4}x - \log_{16}(x+3) = \frac{1}{2}$
    $\displaystyle \log_{16}x= y$ mean $\displaystyle x= 16^y= (4^2)^y= 4^{2y}$ or $\displaystyle 2y= \log_{4}x$ and $\displaystyle y=\log_{16}(x)=\log_4x/2$

    You can write this as $\displaystyle \log_{4}x-\frac{1}{2}\log_4(x+3)= \log_4 x(x+3)^{1/2}= \frac{1}{2}$

    $\displaystyle x(x+3)^{1/2}= 4^{1/2}= 2$.

    Squaring both sides, $\displaystyle x^2(x+ 3)= 4$, $\displaystyle x^3+ 3x^2- 4= 0$. By the "rational root theorem" the only possible rational roots are x= 1, x= -1, x= 2, x= -2, x= 4, and x= -4. Try those to see if any are actually roots.

    [*]$\displaystyle \log(x+12) + \log(x-3) = 2$
    Almost trivial: $\displaystyle \log((x+ 12)(x-3))= 3$. Assuming that "log" here means common logarithm, $\displaystyle (x+12)(x-3)= 10^2= 100$. Solve that quadratic equation.

    [*]$\displaystyle \log_{x^2}9 - \log_{x}36 = 1 $
    Similar to the first one isn't it? $\displaystyle \log_{x^2}9= y$ means $\displaystyle 9= (x^2)^y= x^{2y}$ so $\displaystyle 2y= log_x9$ and $\displaystyle y= \frac{1}{2}log_x9= log_x 9^{1/2}= log_x 3$.

    $\displaystyle \log_x 3- \log_x 36= \log_x \frac{3}{36}= 1$ so $\displaystyle \frac{3}{36}= \frac{1}{12}= x^1$.


    And for this problem, can you check if I solved it right:
    $\displaystyle \log_{x}2 + \log_{x}3 = 5 $
    $\displaystyle \log_{x}(2 * 3) = 5$
    $\displaystyle \log_{x}6 = 5$
    $\displaystyle 6 = x^5$
    $\displaystyle 1.431 = x$
    Yes, that is correct.



    [/quote]Thank you [/QUOTE]
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  3. #3
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    Quote Originally Posted by meiyukichan View Post
    $\displaystyle \log_{4}x - \log_{16}(x+3) = \frac{1}{2}$
    Note that 16 = 4^2. Apply the change-of-base formula to log_16(x) to convert and get:

    . . . . .$\displaystyle 2\log_{16}(x)\, -\, \log_{16}(x\, +\, 3) =\, \frac{1}{2}$

    . . . . .$\displaystyle \log_{16}(x^2)\, -\, \log_{16}(x\, +\, 3)\, =\frac{1}{2}$

    . . . . .$\displaystyle \log_{16}\left(\frac{x^2}{x\, +\, 3}\right)\, =\, \frac{1}{2}$

    Note that log_16(4) = 1/2, so:

    . . . . .$\displaystyle \log_{16}\left(\frac{x^2}{x\, +\, 3}\right)\, =\, \log_{16}(4)$

    Then:

    . . . . .$\displaystyle \frac{x^2}{x\, +\, 3}\, =\, 4$

    Solve the rational equation.

    Quote Originally Posted by meiyukichan View Post
    $\displaystyle \log(x+12) + \log(x-3) = 2$
    A straight-forward application of log rules and the definition of the common log leads to:

    . . . . .$\displaystyle \log\left((x\, +\, 12)(x\, -\, 3)\right)\, =\, \log(100)$

    Equate the arguments, and solve the resulting quadratic equation.

    Quote Originally Posted by meiyukichan View Post
    $\displaystyle \log_{x^2}9 - \log_{x}36 = 1 $
    Note that the change-of-base formula says that log_{x^2}(9) = log_x(9) / log_x(x^2}, and log_x(x^2) obviously equals 2. Then:

    . . . . .$\displaystyle \frac{\log_x(9)}{2}\, -\, \log_x(36)\, =\, \log_x(x)$

    . . . . .$\displaystyle \log_x(\sqrt{9})\, -\, log_x(36)\, =\, \log_x(x)$

    . . . . .$\displaystyle \log_x\left(\frac{3}{36}\right)\, =\, \log_x(x)$

    Equate the arguments, and simplify to solve.

    Quote Originally Posted by meiyukichan View Post
    And for this problem, can you check if I solved it right:

    $\displaystyle \log_{x}2 + \log_{x}3 = 5 $
    $\displaystyle \log_{x}(2 * 3) = 5$
    $\displaystyle \log_{x}6 = 5$
    $\displaystyle 6 = x^5$
    $\displaystyle 1.431 = x$
    Looks good to me! :wink:
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