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Thread: Quick algebra question (logs)

  1. #1
    Junior Member reynardin's Avatar
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    Quick algebra question (logs)

    How do I evaluate this expression?
    z^log k
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    How do I evaluate this expression?
    z^log k
    Is that $\displaystyle z^{\log_{10}(k)}$?

    If so The only way I can see that you can simplify this.

    $\displaystyle z^{log_{10}(k)}=z^{\frac{\log_z(k)}{log_z(10)}}=\b igg(z^{\log_z(k)}\bigg)^{\log_z(10)}=k^{\log_z(10) }$

    I cannot really see much else
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  3. #3
    Junior Member reynardin's Avatar
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    ah, i figured out why i was having trouble. it was z^log base z of k, with the z in microprint..
    Last edited by ThePerfectHacker; May 21st 2008 at 12:37 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    ah, i figured out why i was having trouble. it was z^log base z of k, with the z in microprint..
    Ok..just in case though

    $\displaystyle z^{\log_z(k)}=k$

    This is due to the fact that $\displaystyle \log_z(u(x))$ and $\displaystyle z^{u(x)}$ are inverses

    and by definition $\displaystyle f(f^{-1}(u(x))=u(x)$

    This is true in this case but not always

    For example consider $\displaystyle x^2$ and $\displaystyle \sqrt{x}$
    Last edited by ThePerfectHacker; May 21st 2008 at 12:38 PM.
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  5. #5
    Junior Member reynardin's Avatar
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    Ok, another math question:
    using the values $\displaystyle \log_3(5)=1.4650
    $ and $\displaystyle \log_3(2)=.6309$, how would I solve $\displaystyle log_3(45)$

    I get as far as $\displaystyle \log_3(5)+\log_3(9)$, but I don't know what to do next.

    heck, how do I use the give value of a log with something like $\displaystyle log_3(1.2)$
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    Ok, another math question:
    using the values $\displaystyle \log_3(5)=1.4650
    $ and $\displaystyle \log_3(2)=.6309$, how would I solve $\displaystyle log_3(45)$

    I get as far as $\displaystyle \log_3(5)+\log_3(9)$, but I don't know what to do next.

    heck, how do I use the give value of a log with something like $\displaystyle log_3(1.2)$
    Ok you were good as you were

    but you must see that $\displaystyle \log_3(5)+\log_3(9)=\log_3(5)+\log_3(3^2)=$$\displaystyle \log_3(5)+2\log(3)=1.4560+2$

    and for the second one

    $\displaystyle \log_3(1.2)=\log_3\bigg(\frac{6}{5}\bigg)=\log_3(6 )-\log_3(5)=$$\displaystyle \log_3(3\cdot{2})-\log_3(5)=\log_3(3)+\log_3(2)-\log_3(5)=1+.6309-1.4560=.1749$
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  7. #7
    Junior Member reynardin's Avatar
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    thanks for being such a huge help =) I'm actually starting to get this stuff now. Still got a bit to go though... about 9 problems left that I don't understand. Can I hit them off you?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    thanks for being such a huge help =) I'm actually starting to get this stuff now. Still got a bit to go though... about 9 problems left that I don't understand. Can I hit them off you?
    Of course...I wouldnt come on here if I wasnt in the mood to be "hit off of"
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  9. #9
    Junior Member reynardin's Avatar
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    thanks!
    ok,
    $\displaystyle 1/6=6^{n+4}$
    I know the answer is n=-5, but I don't know how to do the intermediate steps.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    thanks!
    ok,
    $\displaystyle 1/6=6^{n+4}$
    I know the answer is n=-5, but I don't know how to do the intermediate steps.
    The key to these is get the bases the same

    So rewriting we get $\displaystyle \frac{1}{6}=6^{-1}=6^{n+4}$

    taking the $\displaystyle \log_6$ of both sides we get

    $\displaystyle -1=n+5\Rightarrow{n=-5}$
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  11. #11
    Junior Member reynardin's Avatar
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    Ah.. i get it now. 1 down, 8 to go =)

    $\displaystyle \log_5(X)=\bigg(\frac{1}{2}\bigg)\log_5(25)$

    If I understand correctly, should that turn into
    $\displaystyle \log_5(X)=\log_5(25^\frac{1}{2})$?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    Ah.. i get it now. 1 down, 8 to go =)

    $\displaystyle \log_5(X)=\bigg(\frac{1}{2}\bigg)\log_5(25)$

    If I understand correctly, should that turn into
    $\displaystyle \log_5(X)=\log_5(25^\frac{1}{2})$?
    Yes exactly

    So using that we get $\displaystyle \log_5(x)=\log_5(\sqrt{25})=\log_5(5)=1$

    So now we see that $\displaystyle \log_5(5)=1$

    There are two ways of looking at this...the direct way

    $\displaystyle 5^{\log_5(x)}=5^1\Rightarrow{x=5}$

    or also noting that the only number that makes $\displaystyle \log_b$ equal to one is b...so since we have $\displaystyle \log_5(x)=1$ the only value that makes this true is the base value which is 5
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  13. #13
    Junior Member reynardin's Avatar
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    yay!

    Ok, for this one though, I don't really know where to start:
    $\displaystyle \log_2(2X+6)-\log_2(X)=3$

    Just give me enough to get me started, I don't want the whole problem done for me =)
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    yay!

    Ok, for this one though, I don't really know where to start:
    $\displaystyle \log_2(2X+6)-\log_2(X)=3$

    Just give me enough to get me started, I don't want the whole problem done for me =)
    Good! That is the spirit!

    Consider $\displaystyle \log_2(2x+6)-\log_2(x)=\log_2\bigg(\frac{2x+6}{x}\bigg)$
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  15. #15
    Junior Member reynardin's Avatar
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    hm, from there I get to $\displaystyle 2^2 = \frac{6}{x}$ Good so far?
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