Results 1 to 11 of 11

Math Help - help me to solve the equations

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    21

    help me to solve the equations

    i want to find A,B & C from the equation below,

    100 = (4A+2B)*EXP^-(2c)
    120 = (16A+4B)*EXP^-(4c)
    130 = (36A+6B)*EXP^-(6c)

    help me to find these coefficients (A,B,C)..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2008
    Posts
    73
    If its negative exponent we have x^-1 = 1/x

    So you could rewrite the problems with denominators. Then you would have something like

    100 = 2C/(4A + 2B)

    Then multiply the whole equation by the denominator to get rid of the denominators. Then solve for one of the variables and substitute into another equation and keep going until you find all the variables.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    21
    thanks caity for the reply



    but i didnt know how you got the above equation. please explain me.

    (Then multiply the whole equation by the denominator to get rid of the denominators)

    i didnt understand how to do this for this equation too.

    please clear me guys....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2008
    Posts
    73
    I used the formula x^-1 = 1/x

    so I let the -2C be the -1 and 4A + 2B be the x but I tried to work this problem for you and ended up with 0's each time. I think that step is right but think need to solve it differently from there and I'm not sure how. I'll play with it some more or maybe someone else can help us out here. You could also try guess and check when its in the fraction form because it would be easier to work with.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    21
    thansk caity

    i think the equations are not clear it is
    100 = (4A+2B) * e^(-2C) = (4A+2B) / e^(2C)
    120 = (16A+4B) * e^(-4C) = (16A+4B) / e^(4C)
    130 = (36A+6B) * e^(-6C) = (36A+6B) / e^(66C)

    ya i also tried the ways by keeping them to denominator, but it will make A in terms of B and either ways tats the problem. took logarthimic methods also that was also not coming good, and expansion also same, i dont know how to play with this equation.

    when we divide these equations it leads to a fraction in LHS.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2008
    Posts
    73
    Ok I see now... I suggest try sending one of the big guys a pm and see if they can help... sorry I didnt see the e in the beginning I though you just meant exponent.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    1

    Business Math, need help!

    I am stuck on the following problem.

    The list price for a radio is 22% higher than its net price. If the net price is $29.00, what is the list price? What is the amount of the trade discount? HELP!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    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 wgazell View Post
    I am stuck on the following problem.

    The list price for a radio is 22% higher than its net price. If the net price is $29.00, what is the list price? What is the amount of the trade discount? HELP!!!
    Hey wgazell! Don't hijack someone else's thread. Make one of your own.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, sanmo4!

    I had a good start on it . . . then got bogged down.



    Find a, b\text{ and }c\!:\;\;\begin{array}{cccc} 100 &=&(4a + 2b) e^{-2c} \\ 120 &=& (16a + 4b)e^{-4c} \\130 &=& (36a + 6b)e^{-6c} \end{array}
    The equations can be simplified . . .

    . . \begin{array}{cccccccc}<br />
100 &=& \dfrac{2(2a+b)}{e^{2c}} & \Longrightarrow & e^{2c} &=& \dfrac{2a+b}{50} & {\color{blue}[1]} \\ \\[-3mm]<br /> <br />
120 & = & \dfrac{4(4a+b)}{e^{4c}} & \Longrightarrow & e^{4c} &=& \dfrac{4a+b}{30} & {\color{blue}[2]} \\ \\[-3mm]<br /> <br />
130 &=& \dfrac{6(6a+b)}{e^{6c}} & \Longrightarrow & e^{6c} &=& \dfrac{3(6a+b)}{65} & {\color{blue}[3]} \end{array}


    \begin{array}{cccc}\text{Square [1]:} & e^{4c} &=& \dfrac{(2a+b)^2}{50^2} \\ \\[-3mm] \text{Equate to [2]:} & e^{4c} &=& \dfrac{4a+b}{30} \end{array}\qquad\Rightarrow\qquad \frac{(2a+b)^2}{50^2} \:=\:\frac{4a+b}{30}\;\;{\color{blue}[4]}


    \begin{array}{cccc} \text{Multiply [1] and [2]:} & e^{6c} &=&\dfrac{(2a+b)(4a+b)}{1500} \\ \\[-3mm] \text{Equate to [3]:} & e^{6c} &=& \dfrac{3(6a+b)}{65}\end{array} .  \Rightarrow\qquad \frac{(2a+b)(4a+b)}{1500} \:=\:\frac{3(6a+b)}{65}\;\;{\color{blue}[5]}



    I thought that, between [4] and [5], I could solve for a\text{ and }b, but . . .

    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Sep 2008
    Posts
    21
    thanks sorobon and caity...

    in the previous post, if we equate eqns 4 & 5 we get a in terms of b and viceversa.

    i have a approacing in another way but i am not sure whether it is correct or not, please clarify me.
    my approach is below..

    2a + b = 50e^2c ....... 1
    4a + b = 30e^4c ....... 2
    18a + 3b = 65e^6c ....... 3

    from 1, b = 50e^2c - 2a

    substitute for b in eqn 2, we get.

    4a + 50e^2c - 2a = 30e^4c

    2a = 30e^4c - 50e^2c
    a = ( 30e^4c - 50e^2c )/2

    substitute for b, we get..

    b = 50e^2c - 30e^4c - 50e^2c
    b = 100e^2c - 30e^4c

    substitute for a & b in eqn 3.. we get,

    18 * ( 30e^4c - 50e^2c )/2 + 3* 100e^2c - 30e^4c = 65e^6c
    270e^4c - 450e^2c + 300e^2c - 90e^4c = 65e^6c
    180e^4c - 150e^2c = 65e^6c

    divide by e^2c, we get

    65e^4c -180e^2c + 150 = 0

    assume e^2c = x, therefore eqn bcomes

    65x^2 -180x +150 =0

    by solving the quadratic equation we can get c, a & b can be found subsequently.

    but i tried two problems in this form but in both it comes solution doesnt exixt because it leads to sqrt( negative no)...

    please clarify me whether the approach is correct...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Sep 2008
    Posts
    21

    Smile

    hi guys,

    i thank everyone for your replies.... i back substitued the answer in the formula, the formula and solution is correct.. thankyou for your ideas....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. equations solve
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 30th 2011, 06:46 AM
  2. Is it possible to solve these equations?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 8th 2011, 02:43 PM
  3. Solve the following equations for 0<x<360 ?
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: January 31st 2011, 12:33 AM
  4. Equations - Can't solve it!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 22nd 2009, 02:22 AM
  5. solve equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 9th 2008, 07:04 PM

Search Tags


/mathhelpforum @mathhelpforum