help me to solve the equations

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October 15th 2008, 04:49 AM
sanmo4

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)..
October 15th 2008, 07:30 AM
Caity

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.

(Dance)
October 15th 2008, 08:18 PM
sanmo4

thanks caity for the reply

http://www.mathhelpforum.com/math-he...f72c1c3f-1.gif

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....
October 16th 2008, 09:06 AM
Caity

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.
October 16th 2008, 07:56 PM
sanmo4

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.
October 17th 2008, 05:34 AM
Caity

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.
October 17th 2008, 07:04 AM
wgazell

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!!!
October 17th 2008, 07:38 AM
masters

Quote:

Originally Posted by wgazell

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.
October 17th 2008, 02:04 PM
Soroban

Hello, sanmo4!

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

Quote:

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 . . .

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

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

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

I thought that, between [4] and [5], I could solve for $a\text{ and }b$ , but . . .
October 17th 2008, 11:00 PM
sanmo4

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...
October 23rd 2008, 10:26 PM
sanmo4

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....