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Thread: properties of logarithms

  1. #1
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    properties of logarithms

    I have the expression $\displaystyle 4log_{3}\sqrt[2]{x}-4log_{3}x$. In simplifying the first term I need to get rid of the square root and change the coefficient of 4 to an exponent. So if I get rid of the square root first that would be $\displaystyle x^{1/2}$ so when I now change the coefficient of 4 to an exponent, shouldnt the term be $\displaystyle log_{3}(x^{1/2})^4$? It seems that the actual answer is $\displaystyle log_{3}x^{1/2(4)}$ but according to the properties of logarithms I dont understand why. The properties say that $\displaystyle \sqrt[x]{y}=y^{1/x}$ and $\displaystyle mlog n=logn^m$ so why is the simplification of the term $\displaystyle log_{3}x^{1/2(4)}$ and not $\displaystyle log_{3}(x^{1/2})^4$?

    In other words if $\displaystyle mlog n=logn^m$ then why am I simply multiplying the exponent of 1/2 in $\displaystyle x^{1/2}$ by 4 instead of raising it to a power of 4?
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  2. #2
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    Quote Originally Posted by mgrade2000 View Post
    I have the expression $\displaystyle 4log_{3}\sqrt[2]{x}-4log_{3}x$. In simplifying the first term I need to get rid of the square root and change the coefficient of 4 to an exponent. So if I get rid of the square root first that would be $\displaystyle x^{1/2}$ so when I now change the coefficient of 4 to an exponent, shouldnt the term be $\displaystyle log_{3}(x^{1/2})^4$? It seems that the actual answer is $\displaystyle log_{3}x^{1/2(4)}$ but according to the properties of logarithms I dont understand why. The properties say that $\displaystyle \sqrt[x]{y}=y^{1/x}$ and $\displaystyle mlog n=logn^m$ so why is the simplification of the term $\displaystyle log_{3}x^{1/2(4)}$ and not $\displaystyle log_{3}(x^{1/2})^4$?

    In other words if $\displaystyle mlog n=logn^m$ then why am I simply multiplying the exponent of 1/2 in $\displaystyle x^{1/2}$ by 4 instead of raising it to a power of 4?
    Let $\displaystyle x>0$ and $\displaystyle a,b \in R$ , then :
    $\displaystyle a log (x)^b = log \left( (x)^b \right)^a = log (x)^{ab}$.

    You should know that $\displaystyle (a^b)^c=a^{bc}$.
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  3. #3
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    $\displaystyle log_{3}x^{(1/2)4}=log_{3}x^{4/2}=log_{3}x^{2}$
    i think this could help you...i'm not sure if am right but kind of having trouble in logs
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  4. #4
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    Hello mgrade2000

    You're asking lots of questions here. Let me see if I can answer them one at a time.
    Quote Originally Posted by mgrade2000 View Post
    I have the expression $\displaystyle 4log_{3}\sqrt[2]{x}-4log_{3}x$. In simplifying the first term I need to get rid of the square root and change the coefficient of 4 to an exponent. So if I get rid of the square root first that would be $\displaystyle x^{1/2}$
    Correct!
    so when I now change the coefficient of 4 to an exponent, shouldnt the term be $\displaystyle log_{3}(x^{1/2})^4$?
    Yes. It is.
    It seems that the actual answer is $\displaystyle log_{3}x^{1/2(4)}$ but according to the properties of logarithms I dont understand why.
    I think that at the heart of your confusion is that you're not sure what $\displaystyle (x^{\frac12})^4$ means. It means:
    $\displaystyle (x^{\frac12})\times(x^{\frac12})\times(x^{\frac12} )\times(x^{\frac12})$
    which is the same as:
    $\displaystyle x^{(\frac12+\frac12+\frac12+\frac12)}=x^{(\frac12\ times4)}$
    Notice that it isn't:
    $\displaystyle x^{(\frac12)^4}$
    That would be:
    $\displaystyle x^{(\frac12\times\frac12\times\frac12\times\frac12 )}=x^{(\frac{1}{16})}$
    which is something different altogether.

    The properties say that $\displaystyle \sqrt[x]{y}=y^{1/x}$ and $\displaystyle mlog n=logn^m$ so why is the simplification of the term $\displaystyle log_{3}x^{1/2(4)}$ and not $\displaystyle log_{3}(x^{1/2})^4$?
    Can you see now that these are the same thing?
    In other words if $\displaystyle mlog n=logn^m$ then why am I simply multiplying the exponent of 1/2 in $\displaystyle x^{1/2}$ by 4 instead of raising it to a power of 4?
    See above!

    Grandad
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  5. #5
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    yes it makes sense now. Thanks a lot.
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