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)..
If its negative exponent we have x^-1 = 1/x
So you could rewrite the problems with denominators. Then you would have something like
$\displaystyle 100 = 2C/(4A + 2B)$
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.
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.
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.
Hello, sanmo4!
I had a good start on it . . . then got bogged down.
The equations can be simplified . . .Find $\displaystyle a, b\text{ and }c\!:\;\;\begin{array}{cccc} 100 &=&(4a + 2b) e^{-2c} \\ 120 &=& (16a + 4b)e^{-4c} \\130 &=& (36a + 6b)e^{-6c} \end{array}$
. . $\displaystyle \begin{array}{cccccccc}
100 &=& \dfrac{2(2a+b)}{e^{2c}} & \Longrightarrow & e^{2c} &=& \dfrac{2a+b}{50} & {\color{blue}[1]} \\ \\[-3mm]
120 & = & \dfrac{4(4a+b)}{e^{4c}} & \Longrightarrow & e^{4c} &=& \dfrac{4a+b}{30} & {\color{blue}[2]} \\ \\[-3mm]
130 &=& \dfrac{6(6a+b)}{e^{6c}} & \Longrightarrow & e^{6c} &=& \dfrac{3(6a+b)}{65} & {\color{blue}[3]} \end{array}$
$\displaystyle \begin{array}{cccc}\text{Square [1]:} & e^{4c} &=& \dfrac{(2a+b)^2}{50^2} \\ \\[-3mm] \text{Equate to [2]:} & e^{4c} &=& \dfrac{4a+b}{30} \end{array}\qquad\Rightarrow\qquad \frac{(2a+b)^2}{50^2} \:=\:\frac{4a+b}{30}\;\;{\color{blue}[4]}$
$\displaystyle \begin{array}{cccc} \text{Multiply [1] and [2]:} & e^{6c} &=&\dfrac{(2a+b)(4a+b)}{1500} \\ \\[-3mm] \text{Equate to [3]:} & e^{6c} &=& \dfrac{3(6a+b)}{65}\end{array}$ . $\displaystyle \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 $\displaystyle a\text{ and }b$, but . . .
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...