How to solve it ?

last SG · 2009-06-20T23:17:12+00:00

I don't have a calculator: x lg 2.3 + (1-x) lg 3 = lg 11 x lg 2.3 + 1 lg 3 - x lg 3 = lg 11 x lg 2.3 - x lg 3 = lg 11 - 1 lg 3 x (lg 2.3 - lg 3) = lg 11 - 1...

ThunderFbolt SG

log (a + b) =/= log a + log b

log (ab) then is log a + log b

looking for someone to do this correctly though, can't do it either =P

deepak.c SG

Somebody who claims can't do simple maths is now expert at logarithm.

*Looking in the general direction of charlize*

charlize SG

I did some googling.

Apparently log (a+b) is just log (a+b).

deepak.c SG

Can google for me how to build a Faraday Cage?

I can't seem to find directions on how to construct one.

Lee012lee SG

Solve the equation

2.3^^x + 3^^(1-x) = 11

Answer : x = 2.865

Thanks for the help.

If this a H2 Maths question, we can use a graphic calculator

eg TI-84 Plus to solve it.

Steps

(1) Plot the curve y = 2.3^x

(2) Plot the curve y = 11 - 3^(1 - x)

(3) Find the x-ordinate of the intersecting point of the 2 curves

(a) press 2nd Calc

(b) choose 5 for intersect

(4) x = 2.8647872

gratitude SG

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)

ThunderFbolt SG

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.

ThunderFbolt SG

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 =)

Lee012lee SG

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 .

deepak.c SG

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.

charlize SG

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?

Forbiddensinner SG

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.

Puppey SG

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

Lee012lee SG

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

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)

Uncertain SG

Lee012lee is right. Only Newton's Raphson method and other approximation method can solve this problem.

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