Results 1 to 12 of 12

Math Help - Are the Logs or ln logs

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Are the Logs or ln logs

    I have a log question and was wondering which is the correct way to find the value of;

    -x being part of the exponent.

    log 4^3 - x = log 7

    I thought I could find the value of -x this way;

    log4^3-x = log7

    bring down the -x

    xlog64 = log7

    divide both sides by log64

    x = log7 divided by log64

    x = 0.47

    Is this permissible?

    Thanks

    David
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Are the Logs or ln logs

    I guess the equation is:
    \log(4^{3-x})=\log(7)
    You can do the following step:
    (3-x)\log(4)=\log7
    \Leftrightarrow ...
    Or you can solve it like:
    \log(4^{3-x})=\log(7)
    \Leftrightarrow \log(4^3\cdot 4^{-x})=\log(7)
    \Leftrightarrow \log(4^3)+\log(4^{-x})=\log(7)
    \Leftrightarrow ...

    Do you notice the difference with your solution (which is incorrect)? You can't just put -x before the log without doing the same for the 3, you can put (3-x) before the logarithm or you can split the power.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Are the Logs or ln logs

    Quote Originally Posted by Siron View Post
    I guess the equation is:
    \log(4^{3-x})=\log(7)
    You can do the following step:
    (3-x)\log(4)=\log7
    \Leftrightarrow ...
    Or you can solve it like:
    \log(4^{3-x})=\log(7)
    \Leftrightarrow \log(4^3\cdot 4^{-x})=\log(7)
    \Leftrightarrow \log(4^3)+\log(4^{-x})=\log(7)
    \Leftrightarrow ...

    Do you notice the difference with your solution (which is incorrect)? You can't just put -x before the log without doing the same for the 3, you can put (3-x) before the logarithm or you can split the power.

    Yes I have noticed that reading through your thread.

    Can I ask this because I have seen many posters use the DOUBLE SIDED ARROW in your post, does that mean SAME AS?

    David
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Are the Logs or ln logs

    It means equivalent to ... take for example:
    \ln(e)=1 \Leftrightarrow e^{1}=e
    It's an equivalence, because
    If \ln(e)=1 then e^{1}=e, worded in mathematical symbols: \ln(e)=1 \Rightarrow e^1=e, but also
    If e^1=e then \ln(e)=1, worded in mathematical symbols: \ln(e)=1 \Leftarrow e^1=e
    So:
    \ln(e)=1 \Leftrightarrow e^1=e

    But can you solve the equation now? What's your solution now?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Are the Logs or ln logs

    Quote Originally Posted by Siron View Post
    It means equivalent to ... take for example:
    \ln(e)=1 \Leftrightarrow e^{1}=e
    It's an equivalence, because
    If \ln(e)=1 then e^{1}=e, worded in mathematical symbols: \ln(e)=1 \Rightarrow e^1=e, but also
    If e^1=e then \ln(e)=1, worded in mathematical symbols: \ln(e)=1 \Leftarrow e^1=e
    So:
    \ln(e)=1 \Leftrightarrow e^1=e

    But can you solve the equation now? What's your solution now?
    In a word NO, I can't see now how to seperate the -x from your method as you have combined it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Are the Logs or ln logs

    To solve:
    \log(4^{3-x})=\log(7)
    \Leftrightarrow \log(4^{3}\cdot 4^{-x})=\log(7)
    \Leftrightarrow \log(64)+\log(4^{-x})=\log(7)
    \Leftrightarrow \log(64)-x\log(4)=\log(7)

    Other way:
    \Leftrightarrow (3-x)\log(4)=\log(7)
    \Leftrightarrow 3\log(4)-x\log(4)=\log(7)


    Is this clear?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1

    Re: Are the Logs or ln logs

    Quote Originally Posted by David Green View Post
    In a word NO, I can't see now how to seperate the -x from your method as you have combined it.
    Mr. Green, you are making too much out of this.
    \log(4^{3-x})=(3-x)\log(4)

    (3-x)=\frac{\log(7)}{\log(4)}

    x=\left(3-\frac{\log(7)}{\log(4)}\right)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Are the Logs or ln logs

    Quote Originally Posted by Siron View Post
    To solve:
    \log(4^{3-x})=\log(7)
    \Leftrightarrow \log(4^{3}\cdot 4^{-x})=\log(7)
    \Leftrightarrow \log(64)+\log(4^{-x})=\log(7)
    \Leftrightarrow \log(64)-x\log(4)=\log(7)

    Other way:
    \Leftrightarrow (3-x)\log(4)=\log(7)
    \Leftrightarrow 3\log(4)-x\log(4)=\log(7)


    Is this clear?
    In a word NO, there is an overwhelming amount of information there too much to take in. I think it's best to possibly stick with your second example and then I can learn how to find the value of X from that point onwards.

    Thanks

    David
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Are the Logs or ln logs

    Quote Originally Posted by Plato View Post
    Mr. Green, you are making too much out of this.
    \log(4^{3-x})=(3-x)\log(4)

    (3-x)=\frac{\log(7)}{\log(4)}

    x=\left(3-\frac{\log(7)}{\log(4)}\right)
    Plato, thank you this method is much clearer. Looking back on the other thread had too much detail and I could not get my head round how to seperate the X from the logs, you made this very clear.

    David
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Are the Logs or ln logs

    Quote Originally Posted by David Green View Post
    In a word NO, there is an overwhelming amount of information there too much to take in. I think it's best to possibly stick with your second example and then I can learn how to find the value of X from that point onwards.

    Thanks

    David
    Can you say which step you don't understand etc ... ? (that's important!) Actually it's just basic algebra.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Are the Logs or ln logs

    Quote Originally Posted by Siron View Post
    To solve:
    \log(4^{3-x})=\log(7)
    \Leftrightarrow \log(4^{3}\cdot 4^{-x})=\log(7)
    \Leftrightarrow \log(64)+\log(4^{-x})=\log(7)
    \Leftrightarrow \log(64)-x\log(4)=\log(7)

    Other way:
    \Leftrightarrow (3-x)\log(4)=\log(7)
    \Leftrightarrow 3\log(4)-x\log(4)=\log(7)


    Is this clear?
    Your last thread asked me to explain the part that is not clear?

    If you look at what you posted and try to look at it from a view that somebody doesn't know how to get x on it's own, then look where x is in the logs you could forgive for not knowing how to get it on it's own?

    Looking now at the second example you gave;

    (3 - x)log(4) = log(7), here I would divide the LHS to get log(4) on the RHS

    (3 - x)log(4) / log(4) = log(7) / log(4), this leaves;

    3 - x = log(7) / log(4), from here and only because I have seen it from PLATO thread I know to move the 3 to the RHS and change the sign;

    x = 3 - log(7) / log(4)

    x = 1.596

    x = 1.60 (3.s.f) although I have a little doubt about the 0 because 0 is not normally significant?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Are the Logs or ln logs

    Quote Originally Posted by David Green View Post

    If you look at what you posted and try to look at it from a view that somebody doesn't know how to get x on it's own, then look where x is in the logs you could forgive for not knowing how to get it on it's own?
    Offcourse, but I (and the others) don't know what your knowledge is about algebra so it's difficult to estimate what you already have learned, not learned yet, etc.
    But I think you understand it now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 06:39 PM
  2. Logs
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 24th 2010, 08:52 AM
  3. Logs
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 10th 2009, 07:08 PM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 07:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 09:58 PM

Search Tags


/mathhelpforum @mathhelpforum