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Thread: logarithm problem

  1. #1
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    logarithm problem

    Show that for any positive numbers $\displaystyle x$ and $\displaystyle a$ (with $\displaystyle x$ not equal to $\displaystyle 1$) $\displaystyle \log_{x^2} a = \frac{1}{2} \log_{x} a $.

    Also, generalize the assertion in the problem above to powers other than $\displaystyle x^2 $.




    Thank you (:
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  2. #2
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    Hi.


    Quote Originally Posted by meiyukichan View Post
    Show that for any positive numbers $\displaystyle x$ and $\displaystyle a$ (with $\displaystyle x$ not equal to $\displaystyle 1$) $\displaystyle \log_{x^2} a = \frac{1}{2} \log_{x} a $.

    Also, generalize the assertion in the problem above to powers other than $\displaystyle x^2 $.

    Do you know this rule: $\displaystyle log_(y)x = \frac{ln(x)}{ln(y)}$ (rule1)?

    In this case $\displaystyle \log_{x^2} a = \frac{ln(a)}{ln(x^2)} = \frac{1}{2}\frac{ln(a)}{ln(x)} $

    because (rule2) $\displaystyle log(x^n) = n log(x)$

    Using (rule1)
    $\displaystyle 1/2 * log_x(a)$

    Do you know these rules or do you have to proove them?
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  3. #3
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    Quote Originally Posted by Rapha View Post
    Hi.





    Do you know this rule: $\displaystyle log_(y)x = \frac{ln(x)}{ln(y)}$ (rule1)?

    In this case $\displaystyle \log_{x^2} a = \frac{ln(a)}{ln(x^2)} = \frac{1}{2}\frac{ln(a)}{ln(x)} $

    because (rule2) $\displaystyle log(x^n) = n log(x)$

    Using (rule1)
    $\displaystyle 1/2 * log_x(a)$

    Do you know these rules or do you have to proove them?
    Now that you mentioned it, I recognize these rules and it helps with the first problem. I understand it now.
    Thank you! (:
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  4. #4
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    The rule follows from:

    If $\displaystyle \log_{x^2} a= y$, then $\displaystyle a= (x^2)^y= x^{2y}$ so $\displaystyle \log_x a= 2y$ and $\displaystyle y= \log_{x^2} a= \frac{1}{2}\log_x a$
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