How to solve it ?
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I can't think of any direct computation method. If you need to solve this equation manually, Newton-Raphson iteration is able to get you to the solution.
Newton-Raphson method:
f(x) = 2.3^x + 3^(1-x) - 11
f'(x) = x*2.3*(x-1) + (1-x)*3^(-x)
x1 = x0 - f(x0) / f'(x0)
Solution:
let x0 = 2: f(x0) = 5.3767, f'(x0) = 4.4889
x1 = 3.1978: f(x1) = 3.1978, f'(x1) = 19.8799
x2 = 2.4325: f(x2) = -3.2086, f'(x2) = 7.9222
x3 = 2.8375: f(x3) = -0.2401, f'(x3) = 13.0293
x4 = 2.8560: f(x4) = -0.0784, f'(x4) = 13.3194
x5 = 2.8618: f(x5) = -0.0262, f'(x5) = 13.4133
x6 = 2.8637: f(x6) = -0.0088, f'(x6) = 13.4446
x7 = 2.8644: f(x7) = -0.0030, f'(x7) = 13.4551
x8 = 2.8647: f(x8) = -0.0010, f'(x8) = 13.4587
x9 = 2.8647
Ans: x = 2.865 (4sf)
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Originally posted by deepak.c:
Can google for me how to build a Faraday Cage?
I can't seem to find directions on how to construct one.
1. buy a hamster and call it Faraday. usually shops sell with cage.
2. throw away hamster
3. you got Faraday's Cage!
to Lee, not sure if that is valid working in an exam though.
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Originally posted by gratitude:
I can't think of any direct computation method. If you need to solve this equation manually, Newton-Raphson iteration is able to get you to the solution.
Newton-Raphson method:
f(x) = 2.3^x + 3^(1-x) - 11
f'(x) = x*2.3*(x-1) + (1-x)*3^(-x)
x1 = x0 - f(x0) / f'(x0)
Solution:
let x0 = 2: f(x0) = 5.3767, f'(x0) = 4.4889
x1 = 3.1978: f(x1) = 3.1978, f'(x1) = 19.8799
x2 = 2.4325: f(x2) = -3.2086, f'(x2) = 7.9222
x3 = 2.8375: f(x3) = -0.2401, f'(x3) = 13.0293
x4 = 2.8560: f(x4) = -0.0784, f'(x4) = 13.3194
x5 = 2.8618: f(x5) = -0.0262, f'(x5) = 13.4133
x6 = 2.8637: f(x6) = -0.0088, f'(x6) = 13.4446
x7 = 2.8644: f(x7) = -0.0030, f'(x7) = 13.4551
x8 = 2.8647: f(x8) = -0.0010, f'(x8) = 13.4587
x9 = 2.8647
Ans: x = 2.865 (4sf)
seems to be the best answer so far =)
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Originally posted by ThunderFbolt:
1. buy a hamster and call it Faraday. usually shops sell with cage.
2. throw away hamster
3. you got Faraday's Cage!
to Lee, not sure if that is valid working in an exam though.
If it is a H2 Maths question, the use of a graphic calculator is allowed in the actual GCE "A" level exam.
I agree that the use of Newton-Raphson method is more appropriate to solve this questions if a graphic calculator cannot be used.
However, it is to be noted that the new syllabus H2 Maths no longer covers the Newton-Raphson method .
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Originally posted by ThunderFbolt:
1. buy a hamster and call it Faraday. usually shops sell with cage.
2. throw away hamster
3. you got Faraday's Cage!
to Lee, not sure if that is valid working in an exam though.
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Originally posted by gratitude:
I can't think of any direct computation method. If you need to solve this equation manually, Newton-Raphson iteration is able to get you to the solution.
Newton-Raphson method:
f(x) = 2.3^x + 3^(1-x) - 11
f'(x) = x*2.3*(x-1) + (1-x)*3^(-x)
x1 = x0 - f(x0) / f'(x0)
Solution:
let x0 = 2: f(x0) = 5.3767, f'(x0) = 4.4889
x1 = 3.1978: f(x1) = 3.1978, f'(x1) = 19.8799
x2 = 2.4325: f(x2) = -3.2086, f'(x2) = 7.9222
x3 = 2.8375: f(x3) = -0.2401, f'(x3) = 13.0293
x4 = 2.8560: f(x4) = -0.0784, f'(x4) = 13.3194
x5 = 2.8618: f(x5) = -0.0262, f'(x5) = 13.4133
x6 = 2.8637: f(x6) = -0.0088, f'(x6) = 13.4446
x7 = 2.8644: f(x7) = -0.0030, f'(x7) = 13.4551
x8 = 2.8647: f(x8) = -0.0010, f'(x8) = 13.4587
x9 = 2.8647
Ans: x = 2.865 (4sf)
Will teacher mark me wrong if I use this method?
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Originally posted by charlize:
Will teacher mark me wrong if I use this method?
I don't think so. But then again, it is dependent on your teacher.
And to MM, touchwood, I wasn't dying on the bed.
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i tried subbing x=2.865 cant get 11 man. i worked it out x=1.50 and 2.3^x + 3^(1-x) = 11
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Originally posted by Puppey:
i tried subbing x=2.865 cant get 11 man. i worked it out x=1.50 and 2.3x + 3(1-x) = 11
Puppey, you have misread the question.
It is 2.3^x + 3^(1-x) = 11.
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Without a graphic calculator, we can still use a scientfic calculator to find the x answer of 2.865 but the method is unlikely to be accepted by teachers.
Question
Solve 2.3^x + 3^(1-x) = 11
Sub x = 2, the answer is less than 11.
So, sub x = 3, the answer is more than 11.
Next, sub x = 2.9,the answer is still more than 11
Sub x = 2.8, the answer is less than 11
Sub x = 2.85, the answer is closer to 11 but still smaller than 11
Sub x = 2.86, the answer is much closer to 11 but still smaller than 11
Sub x = 2.865, the answer is 11.00189664.
So, the x answer is 2.865 (correct to 3 dec. places)
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Lee012lee is right. Only Newton's Raphson method and other approximation method can solve this problem.