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Math Help - Logarithm questions

  1. #1
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    Logarithm questions

    Hi,

    Can anyone help me to solve the following:

    I have to express log[base 10]48 in terms of x, y and z . Given that
    log[10]24= x , log[10]80=y and log[10]25 = z

    Best Regards,
    Ananya
    Last edited by Ananya; March 9th 2009 at 01:49 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ananya View Post
    Hi ,

    I am unable to solve the following questions of log:

    1. If log[html]<sub>10</sub>[/html] 24 = x , log<
    as you can see, the html code is not working for you. please format your questions better in the future.

    what is your question?

    If \log_{10}24 = x, then ...? post the exact instructions
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  3. #3
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    Hi,

    Can anyone help me to solve the following:

    I have to express log[base 10]48 in terms of x, y and z . Given that
    log[10]24= x , log[10]80=y and log[10]25 = z

    Best Regards,
    Ananya
    Last edited by Ananya; March 9th 2009 at 01:48 AM.
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Ananya View Post
    Hi,

    Can anyone help me to solve the following:

    I have to express log[base 10]48 in terms of x, y and z . Given that
    log[10]24= x , log[10]80=y and log[10]25 = z

    Best Regards,
    Ananya
    Hi Ananya



    log_{10}48

    <br />
= log_{10}(24 \times 2) <br />

    =log_{10}(24)+log_{10}(2)

    =log_{10}{(24)+log_{10}(\frac{2^4\times 5}{25^{1/2}})}^{1/4}

    <br />
=x+log_{10}{(\frac{2^4\times 5}{25^{1/2}})}^{1/4}

    <br />
=x+\frac{log_{10}(\frac{2^4\times 5}{25^{1/2}})}{4}

    <br />
=x+\frac{ log_{10} {(2^4\times 5)} - log_{10} 25^{1/2} }{4}

    <br />
=x+\frac{log_{10}{(80)}-z/2}{4}

    <br />
 =x+\frac{y-z/2}{4}

    GO ahead
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  5. #5
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    Smile

    Hi Adarsh,

    Thanks for the solution .

    I have found out another way :

    log[base 10] 48
    =log[base 10] {(24*80*25)/1000}
    = log[10] 24 + log[10] 80 + log[10] 25 - log[10] 1000
    = x + y + z - 3

    Pardon me for posting the same post twice ...i was getting restless and i thought that thread with the same heading can have related posts.

    Best Regards,
    Ananya
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