Results 1 to 7 of 7

Math Help - logarithim

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    4

    logarithim

    Help me prove this logarithmic equation:Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x?
    please help me prove this.. by the way its log base of 3..
    Last edited by PhantomAssasin; June 28th 2010 at 06:13 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Are all of your parentheses correct, there? What exactly is contained in the argument of the logarithm function?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2010
    Posts
    4
    please help me prove that these 2 equations are equal..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Ok, so you need to prove that

    \log_{3}\left(\frac{3x^{2}\cdot 3x}{3x^{2}}\right)=x-x^{2}+2\log_{3}(x).

    Is that correct? I ask because we have to define the problem clearly and carefully before we can do any meaningful work on the problem, and this equation looks very wrong to me.

    The double-click to see LaTeX code feature appears to be back up and running (YAY Admins!). So if you need to learn some LaTeX in order to post your exact problem more clearly, you can double-click on what I've done to get the flavor of what you'd need to do.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2010
    Posts
    4
    tnx for the help.. yeah thats the problem... but its kinda hard to use the code.. sorry if i cant use it now..
    heres my answer:

    Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x

    what i did is i cancel the 3x^2:

    it will become :

    log_3 3x = x - x^2 + 2log_3 x

    then i use this rule: log_a xy = log_a x + log_a y

    so it will become:

    log_3 3 + log_3 x = x - x^2 + 2log_3 x

    since log_3 3 = 1

    then:

    1 + log_3 x = x - x^2 + 2log_3 x

    i multiply both sides by log:

    (1 + log_3 x = x - x^2 + 2log_3 x)log

    the distribute the log:

    log 1 + log(log_3 x) = log x - log x^2 + log(2log_3 x)

    0 + log(log_3 x) = log {[(x)(2log_3 x)]/ x^2}

    i use this rule: log_a x =b ----- x=antilog_a b

    antilog(log(log_3 x) = {[(x)(2log_3 x)]/ x^2}

    then cancel x:

    antilog(log(log_3 x) = (2log_3 x)]/ x

    i use this rule: nlog_a x = log_a x^n

    antilog(log(log_3 x) = (log_3 x^2)]/ x

    use this rule again: log_a x =b ----- x=antilog_a b

    antilog_3{antilog[log(log_3 x)]} = (x^2)/ x

    cancel x:

    antilog_3{antilog[log(log_3 x)]} = x

    use this rule: log_a x =b ----- x=antilog_a b

    log(log_3 x) = log( log_3 x)

    is this right?

    tn for the help!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,701
    Thanks
    1470
    Quote Originally Posted by PhantomAssasin View Post
    tnx for the help.. yeah thats the problem... but its kinda hard to use the code.. sorry if i cant use it now..
    heres my answer:

    Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x

    what i did is i cancel the 3x^2:

    it will become :

    log_3 3x = x - x^2 + 2log_3 x

    then i use this rule: log_a xy = log_a x + log_a y

    so it will become:

    log_3 3 + log_3 x = x - x^2 + 2log_3 x

    since log_3 3 = 1

    then:

    1 + log_3 x = x - x^2 + 2log_3 x
    Yes, so far that is good.

    i multiply both sides by log:

    (1 + log_3 x = x - x^2 + 2log_3 x)log
    This makes no sense at all! "log" is a function, not a number. You cannot "multiply by log"!

    the distribute the log:

    log 1 + log(log_3 x) = log x - log x^2 + log(2log_3 x)
    No, no, no! What you are actually doing is taking the logarithm of each part.
    But log(a+ b) is NOT equal to log(a)+ log(b) so this is incorrect. Multiplication distributes. Functions, in general, do not.

    0 + log(log_3 x) = log {[(x)(2log_3 x)]/ x^2}

    i use this rule: log_a x =b ----- x=antilog_a b
    That's correct (and antilog_a b= a^b) however, since your previous equation was wrong, this is still wrong.

    antilog(log(log_3 x) = {[(x)(2log_3 x)]/ x^2}

    then cancel x:

    antilog(log(log_3 x) = (2log_3 x)]/ x

    i use this rule: nlog_a x = log_a x^n

    antilog(log(log_3 x) = (log_3 x^2)]/ x

    use this rule again: log_a x =b ----- x=antilog_a b

    antilog_3{antilog[log(log_3 x)]} = (x^2)/ x

    cancel x:

    antilog_3{antilog[log(log_3 x)]} = x

    use this rule: log_a x =b ----- x=antilog_a b

    log(log_3 x) = log( log_3 x)

    is this right?

    tn for the help!
    No, it's not, for the reason I stated. Going back to where you were correct,
    1 + log_3 x = x - x^2 + 2log_3 x, subtract log_3 x from both sides and and subtract x- x^2 from both sides.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2010
    Posts
    4
    what do you mean?

    like this:

    1 + log_3 x = x - x^2 + 2log_3 x

    1 + log_3 x - log_3 x -(x - x^2) = x - x^2 + 2log_3 x - log_3 x -(x - x^2)

    x^2 - x + 1 = log_3 x

    then whats next? i dont know what to do...

    and tnx for the help..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. natural logarithim of a single quantitiy
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: July 31st 2010, 06:56 AM
  2. Logarithim log(2x-3)+log(x-4)=1
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 19th 2010, 04:16 PM
  3. logarithim question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 18th 2009, 09:29 AM
  4. Logarithim Question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 16th 2008, 10:55 PM
  5. Logarithim and exponent tutorial
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: April 23rd 2008, 04:13 PM

Search Tags


/mathhelpforum @mathhelpforum