Results 1 to 9 of 9

Math Help - Exponential Equation

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    69

    Exponential Equation

    Hey guys... I am just finishing up my unit, and I had a question that I am stuck on..

    Here it is:

    Solve each of the following exponential equations:

    b) 4(2^x) + 31 (2^x) - 8 = 0

    So, I figure that I can replace (2^x) with just x, and then I would solve it like a complex trinomial.. However, it would then be 4x+31x -8, and I need it to be 4x^2 in order to do that. How am I supposed to solve this?

    THanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hey

    4X+31X-8=(31+4)X-8=35X-8

    Now you can solve the equation 35X-8=0, can't you ? OO
    Why are you looking for a trinomial ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2008
    Posts
    69
    Yeah I can, ... Don't know why I am looking for a trinomial.. I guess cause I was solving the previous questions through factoring. If I subsitute x for (2^x), does that mean I need to stick this value back in at some point?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Slipery View Post
    Hey guys... I am just finishing up my unit, and I had a question that I am stuck on..

    Here it is:

    Solve each of the following exponential equations:

    b) 4(2^x) + 31 (2^x) - 8 = 0

    So, I figure that I can replace (2^x) with just x, and then I would solve it like a complex trinomial.. However, it would then be 4x+31x -8, and I need it to be 4x^2 in order to do that. How am I supposed to solve this?

    THanks
    Hello Slipery,

    Why even replace 2^x?

    Just add the 8 to both sides, factor out the 2^x, and then take the log of everything.

    2^x(4+31)=8

    2^x=\frac{8}{35}

    \ln 2^x = \ln \frac{8}{35}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2008
    Posts
    69
    Thanks guys!

    Masters, I am using your method, and I know how to solve it..

    When I am writing it down (just asking cause this is a question that will be marked), I just solve this by changing



    to

    http://s180.photobucket.com/albums/x...=logquest3.jpg

    I am sorry, I do have a rather basic grasp of logs... I am just wondering, how would I show my work to completion using your method of



    Because I would just transform this into the diagram I showed above
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Slipery View Post
    Thanks guys!

    Masters, I am using your method, and I know how to solve it..

    When I am writing it down (just asking cause this is a question that will be marked), I just solve this by changing



    to

    logquest3.jpg - Image - Photobucket - Video and Image Hosting

    I am sorry, I do have a rather basic grasp of logs... I am just wondering, how would I show my work to completion using your method of



    Because I would just transform this into the diagram I showed above
    Hello Slipery,

    Review your logarithm rules.

     \ln 2^x = \ln \frac{8}{35}

    x \ln 2= \ln 8 - \ln 35

    x=\frac{\ln 8 - \ln 35}{\ln 2}

    x=
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2008
    Posts
    69
    Ahhhh, yes.. I knew that, but I always have a problem remembering when how and which rules to apply =( Thanks alot though
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by Slipery View Post
    Hey guys... I am just finishing up my unit, and I had a question that I am stuck on..

    Here it is:

    Solve each of the following exponential equations:

    b) 4(2^x) + 31 (2^x) - 8 = 0

    So, I figure that I can replace (2^x) with just x, and then I would solve it like a complex trinomial.. However, it would then be 4x+31x -8, and I need it to be 4x^2 in order to do that. How am I supposed to solve this?

    THanks
    I don't want to pick at you, but could it be that the original equation reads:

    4 \cdot 2^{2x} + 31\cdot 2^x-8=0

    If and only if: You have a quadratic equation in 2^x. Use the substitution y = 2^x. Your equation becomes:

    4y^2+31y-8=0~\implies~ y = \dfrac14~\vee~y=-8

    Now solve

    2^x=\dfrac14 = 2^{-2}~\implies~x = -2 or

    2^x = -8~\implies~x\notin \mathbb{R}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2008
    Posts
    69
    The original question was correct. You are also correct earboth.. I just didn't know when and when not to use the substitution. The one and only example for the 'substitution' (my book is very limited with examples, and the explaining of the examples), looks the way that your question looked earboth'
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponential Equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 14th 2010, 01:34 AM
  2. Exponential Equation
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 12th 2009, 11:19 AM
  3. exponential equation
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 24th 2009, 06:38 AM
  4. Exponential equation
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 9th 2009, 08:23 AM

Search Tags


/mathhelpforum @mathhelpforum