Results 1 to 10 of 10

Math Help - ln and ex log questions

  1. #1
    Member
    Joined
    Oct 2006
    From
    Toronto
    Posts
    86

    ln and ex log questions

    Here are some ln (lawn) and e^x questions. I'm still learning these so I'm not so good at solving thema and these ones are hard so I was wondering if anyone can help.



    If that picture doesn't work then here's a link
    http://img329.imageshack.us/img329/3...p141922wy7.png
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie Hootenanny's Avatar
    Joined
    Dec 2006
    Posts
    1
    How much do you know about logarithms? To me these questions appear to be elementary applications of the laws of logs. Could you be more specific as to where you're having problems?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Solve e^x-3e^{-x}=2

    Multiply through by e^x and rearrange to get:

    e^{2x}-2e^x-3=0

    Now put u=e^x and you get a quadratic:

    u^2-2u-3=0.

    Solve this and then the solutions for x are the natural logarithms of the u's.

    The seventh question - Solve e^x+7e^{-x}=8
    can be solved by the same method.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, SportfreundeKeaneKent!

    Here are a couple of them . . .


    2) Solve: . (\ln x)^2\:=\:\ln(x^2)

    We have: . (\ln x)^2\:=\:2\ln x\quad\Rightarrow\quad (\ln x)^2 - 2\ln x \:=\:0

    Factor: . \ln x(\ln x - 2)\:=\:0


    And we have two equations to solve:

    . . \ln x \,= \,0\quad\Rightarrow\quad\boxed{ x \,= \,1}

    . . \ln x \,=\,2\quad\Rightarrow\quad\boxed{ x \,= \,e^2}



    3) The graph of y = \ln x is rotated 90 CCW about the origin.
    What is the equation of the new graph?

    The graph of y = \ln x looks like this:
    Code:
                |
                |                *
                |           *
          ------+-------*----------
                |    *  1
                |  *
                | *
                |
                |*
                |

    Rotated 90 CCW, it looks like this:
    Code:
               *      |
                      |
                *     |
                 *    |
                   *  |
                      *
                      |   *
                      |         *
                      |                 *
          ------------+--------------------
                      |

    We're expected to recognize this as the graph of: y \:=\:e^{-x}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2006
    From
    Toronto
    Posts
    86
    We're sort of doing the log unit at school right now. I thought that these seemed pretty simple too but there's always one extra thing you have to do on some of these harder problems that the teacher gives. The ones from the textbook aren't a problem. lnx is a bit more of a problem and I'm still learning that.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,805
    Thanks
    116
    Quote Originally Posted by SportfreundeKeaneKent View Post
    We're sort of doing the log unit at school right now. I thought that these seemed pretty simple too but there's always one extra thing you have to do on some of these harder problems that the teacher gives. The ones from the textbook aren't a problem. lnx is a bit more of a problem and I'm still learning that.
    Hello Sportsfreund,

    I'll send you a .pdf-file to show you how the rules of powers and the rules of logarithm are connected. (It isn't necesary to learn German to understand this file)

    EB
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello again, SportfreundeKeaneKent!

    A couple more . . .


    4) Solve: . \ln\left(\frac{e^2x^2+1}{2x}\right) \:=\:1

    We have: . \frac{e^2x^2+1}{2x} \:=\:e\quad\Rightarrow\quad e^2x^2 + 1 \:=\:2ex\quad\Rightarrow\quad e^2x^2 - 2ex + 1 \:=\:0

    Factor: . (ex - 1)^2\:=\:0\quad\Rightarrow\quad ex - 1\:=\:0\quad\Rightarrow\quad ex \,=\,1\quad\Rightarrow\quad x \,=\,\frac{1}{e}



    6) Solve: . \ln(x^2 + e) - \ln(x + 1)\:=\:1

    We have: . \ln\left(\frac{x^2+e}{x+1}\right) \:=\:1\quad\Rightarrow\quad\frac{x^2+e}{x+1} \:=\:e\quad\Rightarrow\quad x^2 + e \:=\:ex + e\quad\Rightarrow\quad x^2 - ex \:=\:0

    Factor: . x(x - e)\:=\:0


    And we have two equations to solve:

    . . \boxed{x\:=\:0}

    . . x - e \:=\:0\quad\Rightarrow\quad\boxed{x = e}

    and both answers check out . . .

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2006
    From
    Toronto
    Posts
    86
    Could someone check up on these answers for me for these questions:

    For question 1, I got x = 1.098

    For question 5, I got x = 1.946 and x = 0

    For question 2, does x = 0 AND e^x or just e^x/7.389

    And for the last question (e^3x=15), I got x = 0.9026
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2006
    From
    Toronto
    Posts
    86
    Oh and also, I was doing question 4 down that list and I got x = 0.5 which is different than the answer Soroban's answer of 1/e. I was wondering if anyone could check up on this.

    I used the quotient rule in this question so that it became this:
    ln(e^2 x^2+1)-ln2x=1
    2lne^2 x+ 0 - ln2x=1
    2x + 0 - ln2x = 1
    2x(1-ln1)=1
    2x=1
    x-1/2 or 0.5

    When I plugged x = 0.5 into the equation to check my work, it worked out so I'm really confused about wheter what Soroban did was right or the way I've done it is right.
    Last edited by SportfreundeKeaneKent; December 20th 2006 at 02:10 PM. Reason: Not enough information in first one.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,805
    Thanks
    116
    Quote Originally Posted by SportfreundeKeaneKent View Post
    Oh and also, I was doing question 4 down that list and I got x = 0.5 which is different than the answer Soroban's answer of 1/e. I was wondering if anyone could check up on this.

    I used the quotient rule in this question so that it became this:
    ln(e^2 x^2+1)-ln2x=1
    2lne^2 x+ 0 - ln2x=1
    2x + 0 - ln2x = 1
    2x(1-ln1)=1
    2x=1
    x-1/2 or 0.5

    When I plugged x = 0.5 into the equation to check my work, it worked out so I'm really confused about wheter what Soroban did was right or the way I've done it is right.
    Hello Sportsfreund,

    here:

    ln(e^2 x^2+1)-ln2x=1
    2lne^2 x+ 0 - ln2x=1

    you've made a very common mistake: You used the "ln" as a kind of variable and expanded the brackets by rule of distribution. But ln is a function and the contents of the brackets are the arguments of this function.
    You can get rid of the "ln" if you use the following property:

    e^{\ln(x)}=x

    Soroban's solution is 100% correct.

    EB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum