Naturual Logarithms

Neetu · 2010-09-19 11:53:01

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

IS IT OK?

bobbym · 2010-09-19 12:16:11

Hi Neetu;

Yes, I believe so!

Neetu · 2010-09-26 04:19:52

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)=?

bobbym · 2010-09-26 04:36:52

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.

Neetu · 2010-09-26 13:39:32

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..

bobbym · 2010-09-26 19:32:17

Hi Neetu;

Is this what 1) is?

Neetu · 2010-09-27 12:44:32

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)

bobbym · 2010-09-27 13:43:22

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,

Neetu · 2010-09-28 01:30:09

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

bobbym · 2010-09-28 08:30:16

Hi;

It is in equation 2's form.

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

A fraction and a log. They are just switched.

Neetu · 2010-10-01 10:06:21

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)

bobbym · 2010-10-01 16:15:33

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.

DaveRobinsonUK · 2010-10-01 20:29:39

Wow.. 3 pages! What have I started!

Hows things guys? Hope all is well

bobbym · 2010-10-01 20:33:39

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.

Neetu · 2010-10-02 11:58:09

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

Neetu · 2010-10-02 11:59:58

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

bobbym · 2010-10-02 12:18:24

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.

Neetu · 2010-10-03 00:03:09

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

bobbym · 2010-10-03 10:38:35

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?

Neetu · 2010-10-03 10:44:42

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

bobbym · 2010-10-03 10:47:59

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 ?

Neetu · 2010-10-03 10:56:26

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.

bobbym · 2010-10-03 10:59:39

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.

Neetu · 2010-10-03 11:03:12

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

Neetu · 2010-10-03 11:06:10

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)

Math Is Fun Forum

#51 2010-09-19 11:53:01

Re: Naturual Logarithms

#52 2010-09-19 12:16:11

Re: Naturual Logarithms

#53 2010-09-26 04:19:52

Re: Naturual Logarithms

#54 2010-09-26 04:36:52

Re: Naturual Logarithms

#55 2010-09-26 13:39:32

Re: Naturual Logarithms

#56 2010-09-26 19:32:17

Re: Naturual Logarithms

#57 2010-09-27 12:44:32

Re: Naturual Logarithms

#58 2010-09-27 13:43:22

Re: Naturual Logarithms

#59 2010-09-28 01:30:09

Re: Naturual Logarithms

#60 2010-09-28 08:30:16

Re: Naturual Logarithms

#61 2010-10-01 10:06:21

Re: Naturual Logarithms

#62 2010-10-01 16:15:33

Re: Naturual Logarithms

#63 2010-10-01 20:29:39

Re: Naturual Logarithms

#64 2010-10-01 20:33:39

Re: Naturual Logarithms

#65 2010-10-02 11:58:09

Re: Naturual Logarithms

#66 2010-10-02 11:59:58

Re: Naturual Logarithms

#67 2010-10-02 12:18:24

Re: Naturual Logarithms

#68 2010-10-03 00:03:09

Re: Naturual Logarithms

#69 2010-10-03 10:38:35

Re: Naturual Logarithms

#70 2010-10-03 10:44:42

Re: Naturual Logarithms

#71 2010-10-03 10:47:59

Re: Naturual Logarithms

#72 2010-10-03 10:56:26

Re: Naturual Logarithms

#73 2010-10-03 10:59:39

Re: Naturual Logarithms

#74 2010-10-03 11:03:12

Re: Naturual Logarithms

#75 2010-10-03 11:06:10

Re: Naturual Logarithms

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