Results 1 to 15 of 15

Math Help - Logs and their laws

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Logs and their laws

    I have a log question;

    log 9 x = log 5 (base9) + log 6 (base9) - log 6 (base9)

    I have understood it to this point;

    log 9 x = log 9 {30/3}

    log 9 x = log 9 (10)

    I now require to get the "x" on its own, am I going in the right direction from this point as follows;

    x = log 9 (10) = log 9^10 = 1000 000 000

    I am not to sure what to do with the two log 9's but I require to find the value of "x", am I approaching this correctly?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: Logs and their laws

    Quote Originally Posted by David Green View Post
    I have a log question;

    log 9 x = log 5 (base9) + log 6 (base9) - log 6 (base9)

    log 9 x = log 5 (base9)

    ... the last two terms cancel to 0

    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Logs and their laws

    Quote Originally Posted by skeeter View Post
    ...
    I am sorry Skeeter I pushed the wrong key on the last log?

    log9 x = log9 5 + log9 6 - log9 3

    log9 x = log9 {30/3}

    log9 x = log9 (10)

    I require to get "x" on its own but have log9 on each side of the equals?

    x = log 9^10 = 1000 000 000

    Not sure if this is correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: Logs and their laws

    if \log_b{x} = \log_b{y} , then x = y
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Logs and their laws

    I agree but that does not solve for x?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,610
    Thanks
    1574
    Awards
    1

    Re: Logs and their laws

    Quote Originally Posted by David Green View Post
    I agree but that does not solve for x?
    @David Green, why not use basic LaTeX tags?
    [TEX]\log_{9}(5)[/TEX] gives \log_{9}(5).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Logs and their laws

    Quote Originally Posted by Plato View Post
    @David Green, why not use basic LaTeX tags?
    [TEX]\log_{9}(5)[/TEX] gives \log_{9}(5).
    I was told that they are difficult to use and not had any experience with them!

    From above it looks like you are saying that "x" = 5 am I correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Logs and their laws

    Quote Originally Posted by David Green View Post
    I was told that they are difficult to use and not had any experience with them!

    From above it looks like you are saying that "x" = 5 am I correct?
    You have \log_9(10) = \log_9(x) which is the same as x = 10 (see post 4)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Logs and their laws

    Quote Originally Posted by e^(i*pi) View Post
    You have \log_9(10) = \log_9(x) which is the same as x = 10 (see post 4)

    I am still unsure but;

    log (base9) x = Log (base9) 10

    so I need to get "x" on it's own, therefore because I have no practical examples of how this is done, I am doing it right this way;

    Log 10 = 2
    Follow Math Help Forum on Facebook and Google+

  10. #10
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Logs and their laws

    Quote Originally Posted by David Green View Post
    I am still unsure but;

    log (base9) x = Log (base9) 10

    so I need to get "x" on it's own, therefore because I have no practical examples of how this is done, I am doing it right this way;

    Log 10 = 2
    I'm not sure where you get 2 from. If you like you can raise 9 to the power of both sides - this will have the effect of cancelling out the log function
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Logs and their laws

    Quote Originally Posted by e^(i*pi) View Post
    I'm not sure where you get 2 from. If you like you can raise 9 to the power of both sides - this will have the effect of cancelling out the log function
    Log 10 = 2

    This is because the two represents an exponent, therefore 10^2 = 100, although I am not confident because as you say my base is 9, therefore;

    log 10 = 9

    This would be equal to 1 billion as I have posted previously?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Logs and their laws

    Quote Originally Posted by David Green View Post
    Log 10 = 2

    This is because the two represents an exponent, therefore 10^2 = 100, although I am not confident because as you say my base is 9, therefore;

    log 10 = 9

    This would be equal to 1 billion as I have posted previously?
    The definition of a logarithm is: \log_c(b) = a \Leftrightarrow b = c^a

    If we take the expression for a in the first form and put it into the second we get to: b = c^{\log_c(b)}

    Hence if you raise the base of the logarithm to the log expression they "cancel" in the same way that squaring and square root do.

    9^{\log_9(10)} = 9^{\log_9(x)
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Logs and their laws

    Quote Originally Posted by e^(i*pi) View Post
    The definition of a logarithm is: \log_c(b) = a \Leftrightarrow b = c^a

    If we take the expression for a in the first form and put it into the second we get to: b = c^{\log_c(b)}

    Hence if you raise the base of the logarithm to the log expression they "cancel" in the same way that squaring and square root do.

    9^{\log_9(10)} = 9^{\log_9(x)
    I am not sure I follow the above, however;

    log(base9) x = log(base9) 10

    if I subtract log(base9) from both sides;

    log(base9) - log(base9) x = log(base9) - log(base9) 10

    x = 10

    Not sure if this is mathematically correct for logs, but it shows x = 10
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: Logs and their laws

    Quote Originally Posted by David Green View Post
    I am not sure I follow the above, however;

    log(base9) x = log(base9) 10

    if I subtract log(base9) from both sides;

    log(base9) - log(base9) x = log(base9) - log(base9) 10

    x = 10

    Not sure if this is mathematically correct for logs, but it shows x = 10
    Please pay attention to what you are told or people will just think their time is being wasted.

    You have \log_9 (x) = \log_9 (10).

    Post #4 and post #8 tell you excatly what to do. I cannot see why this thread has had to go past post #8.

    Thread closed.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: Logs and their laws

    Quote Originally Posted by Plato View Post
    @David Green, why not use basic LaTeX tags?

    [TEX]\log_{9}(5)[/TEX] gives \log_{9}(5).
    I was told that they are difficult to use and not had any experience with them!
    [snip]
    What rot. Did you take the time to check the veracity of what you were "told"? 15 minutes (less time than has been spent on this thread since post #8) is all it takes to learn how to do most of what you need to do: http://www.mathhelpforum.com/math-help/f47/
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 05:39 PM
  2. Laws of logs
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 4th 2010, 03:52 AM
  3. laws of logs?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 8th 2009, 09:11 AM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 06:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 08:58 PM

Search Tags


/mathhelpforum @mathhelpforum