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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

YES, i HAVE USED THE SAME EQAUTION, GIVEN IN FORUM #50

IS IT OK?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

Yes, I believe so!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

thanks bobbym... can u also tell me wht is the formula to sove ln(x+y)

i mean how cn v open above equation like ln(xy)=ln(x)+ln(y)

wht is ln(x+y)=?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

Let me help with some terminology that may help clear this up. When we say solve we mean an equation. An equation has an equals sign in it.

Ex. y = 2x + 5 is an equation.

Phrases like ln(x+y) are expressions because they have no equals sign.

Although you can do ln(x*y) = ln(x) + ln(y) for certain values of x and y.

ln(x+y) doesn't have anything similar to that. We say it is in lowest terms or it is already simplifed.

There is this identity for it:

Now if you have an equation such as ln(x + y ) = 5 that I can solve by taking the exponential of both sides.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

actually i dont want to solve equation for a specific solution. my problem is different. ok let me try to explain it.

I have this equation.

F(x, y) = y log2(1+ x*z/N*B* y) + a (y 1) ...........(1)

if we do derivative of eq (1) w.r.t y and equate it to 0 then it yields as follows:

ln (x*z / y) - x*z / y = __?___ ..............(2)

i want to find right hand side of equation (2), i know it is some constant but it is exactly equal to what? i want to find that. please help me out.

thanks..

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

Is this what 1) is?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

yes, eq(1) is same like this.

x is also multiplied by z in the numerator, rest is fine

please help me as soon as possible. thanks

*Last edited by Neetu (2010-09-27 12:51:03)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

If:

I get this for the partial derivative with respect to y.

Set that to 0:

Clear the denominator:

Rearrange:

Where ln is the natural log,

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

thanks... i want this equation somewht like equation 2... can u solve it further.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

It is in equation 2's form.

ln (x*z / y) - x*z / y =

A fraction and a log. They are just switched.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

thanks bobbym...

actualy i want exact solution as equation 2

the solution that you have given is like as follows

ln((x*z+BNy)/BNy)-x*z/(BNy+xz)=-aln(2)

we can leave BN according to my project, So now above equation becomes

ln((x*z+y)/y)-x*z/(y+x*z)=-aln(2) ....(3)

Now from above equation (3), i want to derive equation (2), left hand side should be same as in equation(2).

equation(2) is ln (x*z / y) - x*z / y = __?___

please try to solve my problem. thanks!

*Last edited by Neetu (2010-10-01 10:09:52)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

You have several problems here. 3) Is not correct, it is not what I solved for. There is no minus sign in front of the a.

I can rearrange the equation to the form you want but it will not have a constant on the right.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**DaveRobinsonUK****Member**- Registered: 2010-04-24
- Posts: 123

Wow.. 3 pages! What have I started!

Hows things guys? Hope all is well

Can feel it coming together.. Slowly but Surely

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Dave;

Yes, this is all your fault. Everybody knows I can not answer any question in under 400 000 words or posts, whichever is longer.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

hi bobbym... equation (3) is modified from ur equation. I have rearranged it.

you please try to rearrange it to the required form.... if we dnt get constant at right hand side... dunno wry.... u please bring it to the required form by any means.

Thanks

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

In equation (3), Left hand side is also different, that is why i put minus sign at the right hand side

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

Yes, I see that now, sorry, my mistake. I can manipulate it to the form you want but the RHS will no longer just be -a ln(2).

It will have some variables on that side also.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

Hi bobbym...

i dnt wnt variables at RHS. anyways you solve it the best you can. please drive LHS same as required. we can c RHS later on

Thanks

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Neetu;

Do you have any idea as to what values x,y,z are. Are either x or y or z likely to be close to 0?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

y can be 0 or 1. while others are variables. x and z are variable less than 0

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

So you are saying y is a boolean. It can only take two values?

Because that would be illegal.

Or are you saying it is between 0 and 1?

For the x and z are they large like - 34, -321 or small like - .274 ?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

y can only take two values either 0 or 1

Sory in previous post, i hv written wrong. Actualy, x and z are variabes ranging between 0 to 1.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Okay but you see the problem when y is 0

equation(2) is ln (x*z / y) - x*z / y = __?___

That form you require is illegal. See the y in the denominator. You can not divide by 0. y = 1 is fine.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

I also want to know do derivation rules chnages with summation. i mean wd dis symbol ∑

becuase the original equation is

F(x, y) = ∑1 ∑2 y log2(1+ x*z/N*B* y) + a *(∑1 y 1)

∑1 represents summation ranges from p=1 to P

∑2 represents summation ranges from q=1 to Q

x, y and z are functions of p and q

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**Neetu****Member**- Registered: 2010-08-24
- Posts: 35

yes, we are not goin to put values in this equation. that y in the denominator is nt a problem for my project

please help me out wd this equation

*Last edited by Neetu (2010-10-03 11:09:02)*

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