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Math Help - quick log question

  1. #1
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    quick log question


    I was thinking about different log laws and stuff but i don't know how to get to that. It was on a past paper im pratising on at the moment.

    If anyone can help, il really appreciate it.
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  2. #2
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    Logs

    Hello chinkmeista
    Quote Originally Posted by chinkmeista View Post

    I was thinking about different log laws and stuff but i don't know how to get to that. It was on a past paper im pratising on at the moment.

    If anyone can help, il really appreciate it.

    \log_ax^{10} - 2\log_a\left(\frac{x^3}{4}\right)

    = \log_ax^{10} -\log_a\left(\frac{x^3}{4}\right)^2

    = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)

    = \log_a\left(x^{10}\times\frac{16}{x^6}\right)

    = \log_a(16x^4)

    =4\log(2x)

    Grandad

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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello chinkmeista
    \log_ax^{10} - 2\log_a\left(\frac{x^3}{4}\right)

    = \log_ax^{10} -\log_a\left(\frac{x^3}{4}\right)^2

    = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)

    = \log_a\left(x^{10}\times\frac{16}{x^6}\right)

    = \log_a(16x^4)

    =4\log(2x)

    Grandad
    Oh hi, thats fantastic thank you. I kinda understand it, up until like the 3rd step though from:

    = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)


    to


    = \log_a\left(x^{10}\times\frac{16}{x^6}\right)


    how did you flip the 16 and x^6 upside down and merge the x^10 together at the same time? could you explain how you did this bit?

    Again, thanks a lot.
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  4. #4
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    Quote Originally Posted by chinkmeista View Post
    Oh hi, thats fantastic thank you. I kinda understand it, up until like the 3rd step though from:

    = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)


    to


    = \log_a\left(x^{10}\times\frac{16}{x^6}\right)


    how did you flip the 16 and x^6 upside down and merge the x^10 together at the same time? could you explain how you did this bit?

    Again, thanks a lot.
    -\log_x\left(\frac{y}{z}\right) = \log\left(\left(\frac{y}{z}\right)^{-1}\right) = a\log\left(\frac{z}{y}\right)

    You can take the coefficient outside the log and put it into it's power. Anything to the power of -1 is the recipricole.
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  5. #5
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    Quote Originally Posted by chinkmeista View Post
    Oh hi, thats fantastic thank you. [snip]
    No, no .... when I reply it's fantastic. When Grandad replies it's grand ......
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  6. #6
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    Hello chinkmeista
    Quote Originally Posted by chinkmeista View Post
    how did you flip the 16 and x^6 upside down and merge the x^10 together at the same time? could you explain how you did this bit?[/COLOR][/COLOR][/COLOR]

    Again, thanks a lot.
    \log a - \log b = \log (a \div b), and to divide by a fraction, turn it 'upside down' and multiply.

    So \log_a x - \log_a \left(\frac{y}{z} \right) = \log \left(x \times \frac{z}{y} \right ).

    Can you see it OK now?

    Grandad
    Last edited by Grandad; January 9th 2009 at 04:33 AM. Reason: Incomplete reply
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  7. #7
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    Ho-ho

    Quote Originally Posted by mr fantastic View Post
    No, no .... when I reply it's fantastic. When Grandad replies it's grand ......
    Ho! Ho!

    Grandad
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  8. #8
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    Quote Originally Posted by Grandad View Post
    Hello chinkmeista

    \log a - \log b = \log (a \div b), and to divide by a fraction, turn it 'upside down' and multiply.

    So \log_a x - \log_a \left(\frac{y}{z} \right) = \log \left(x \times \frac{z}{y} \right ).

    Can you see it OK now?

    Grandad
    yeah i see it now, thank you very much!
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