Results 1 to 11 of 11

Math Help - Finding domain, solving exponential equation and finding factors.

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    4

    Finding domain, solving exponential equation and finding factors.

    Hi guys! I would really love help with some questions on my exam review. My exam is tomorrow and I'm completely confused! Help would be very much appreciated.


    Find the domain of:
    f(x) = lnX / x^2 + 4


    Solve for X:
    2^x^-^1 = 3^x

    I know I need to make them the same base, but I don't know what would be simplest way to do that here.


    Determine if (x - 1) is a factor of:
    -x^7^6^2 + 950x^3^4^8 - 938x^2^5 - 13

    I have absolutely no clue what to do here, except synthetic division with 762 places...
    Last edited by mr fantastic; December 8th 2011 at 11:48 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Having some issues with exam review

    For the first part, do you know how to find the domain of a function?

    For the next, there are various approaches. What did you try? Here's one suggestion:

    2^{x-1}=3^x

    Taking logs:

    (x-1)\ln{2}=x\ln{3}

    x\ln{2}-\ln{2}=x\ln{3}

    x\ln{2}-x\ln{3}=\ln{2}

    x(\ln{2}-\ln{3})=\ln{2}

    ..etc.

    For the last part, x-1 will be a factor if f(1)=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2011
    Posts
    4

    Re: Having some issues with exam review

    Usually for a rational I would just factor the denominator, but this isn't factorable by the conventional (x +/- a)(x +/- b)... method and I'm not sure if/how the log would affect the answer.

    Ah I see, okay that makes sense.

    How do you know, though? Is there some way of finding that out without using synthetic division?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Having some issues with exam review

    Quote Originally Posted by Oddrick View Post
    Usually for a rational I would just factor the denominator, but this isn't factorable by the conventional (x +/- a)(x +/- b)... method and I'm not sure if/how the log would affect the answer.
    If x^2+4=0, then  x^2=-4. This cannot be the case for real values of x. The only thing restricting the domain is \ln(x).

    Quote Originally Posted by Oddrick View Post
    Ah I see, okay that makes sense.
    I'm glad, although I'd be surprised if you had nothing in your notes to cover this.

    Quote Originally Posted by Oddrick View Post
    How do you know, though? Is there some way of finding that out without using synthetic division?
    Yes, using the factor theorem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2011
    Posts
    4

    Re: Having some issues with exam review

    So since
    X = ln2/ln2-ln3

    would that mean that
    X = 1/-ln3

    or
    X = -ln3


    This is probably a very dumb question. XP
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Having some issues with exam review

    Quote Originally Posted by Oddrick View Post
    So since
    X = ln2/ln2-ln3

    would that mean that
    X = 1/-ln3

    or
    X = -ln3


    This is probably a very dumb question. XP
    Is that how you cancel terms in a fraction?!? Your instructor would be shocked.
    x=\frac{\ln{2}}{\ln{2}-\ln{3}}

    x=\frac{\ln{2}}{\ln{\frac{2}{3}}}

    Now you could use the change of base rule to simplify.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2011
    Posts
    4

    Re: Having some issues with exam review

    Wait, I'm confused as to how you got the
    {\ln{\frac{2}{3}}}
    I missed pretty much all the discussion about logs so this is very foreign to me...

    And for the one about factor theorem, I got a remainder of -2 so it would be no, right? Does that seem right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Having some issues with exam review

    Quote Originally Posted by Oddrick View Post
    Wait, I'm confused as to how you got the
    {\ln{\frac{2}{3}}}
    I missed pretty much all the discussion about logs so this is very foreign to me...

    And for the one about factor theorem, I got a remainder of -2 so it would be no, right? Does that seem right?
    You're right about the factor/remainder theorem.

    However, it's very obvious that you have very little understanding when it comes to the concept of logarithms. You need to catch up on missed work quickly rather than on the night before your exam. You require more help than you can receive in a quick forum session. I suggest finding an online review of this topic, which you'll need to study for a few hours until you have a thorough understanding.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Dec 2011
    Posts
    7

    Re: Having some issues with exam review

    can you explain it

    Find the domain of:
    f(x) = lnX / x^2 + 4
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Having some issues with exam review

    It's supposed to be f(x)=\frac{\ln{x}}{x^2+4}, and as x^2+4\neq{0} for real x, the only thing restricting the domain is \ln(x), which can only take values for x>{0}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Dec 2011
    Posts
    7

    Re: Having some issues with exam review

    thank you =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 27th 2011, 03:28 AM
  2. Need help finding exponential equation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: October 10th 2010, 09:20 PM
  3. Help finding an exponential equation...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 10th 2010, 04:13 PM
  4. Replies: 4
    Last Post: April 23rd 2009, 04:59 AM
  5. Replies: 2
    Last Post: September 23rd 2008, 07:09 PM

Search Tags


/mathhelpforum @mathhelpforum