Results 1 to 8 of 8

Math Help - quick log question

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    16

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    Posts
    16
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    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 ......
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Logs

    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2008
    Posts
    16
    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quick question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 23rd 2009, 08:28 PM
  2. Quick question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 30th 2008, 04:09 PM
  3. Quick question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 13th 2008, 03:51 PM
  4. Quick Question - SPP, CPP, TSP
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 17th 2007, 08:49 PM
  5. Quick's quick question
    Posted in the Number Theory Forum
    Replies: 22
    Last Post: July 9th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum