Solving for NlnN - n1ln(n1) - n2ln(n2) by Adding and Subtracting Terms

[karnten07]

[SOLVED] Logs and rearranging

Homework Statement

I want to show that NlnN - n1ln(n1) - n2ln(n2) is equal to:

-N{(n1/N)ln(n1/N) + (n2/N)ln(n2/N)}

apparently i can do this by the artifice of adding and subtracting the term n1lnN and then regrouping.

Homework Equations

The Attempt at a Solution

I can get the form required but i still have leftover N's and n1s and n2s, anyone know how to do this? thanks

 

[nate808]

Hello, I am a scientist and I would like to help you solve this problem. First, let's start by rewriting the original equation as: NlnN - n1ln(n1) - n2ln(n2) = NlnN - (n1lnN + n2lnN) + (n1lnn1 + n2lnn2)

Now, we can rearrange this equation by factoring out lnN and grouping the remaining terms: NlnN - (n1lnN + n2lnN) + (n1lnn1 + n2lnn2) = lnN(N - n1 - n2) + (n1lnn1 + n2lnn2)

Next, we can use the fact that n1 + n2 = N to substitute into the equation: lnN(N - n1 - n2) + (n1lnn1 + n2lnn2) = lnN(N - (N - n1 - n2)) + (n1lnn1 + n2lnn2)

Simplifying this, we get: lnN(n1 + n2) + (n1lnn1 + n2lnn2) = lnNn1 + lnNn2 + n1lnn1 + n2lnn2

Now, we can use the logarithmic property ln(ab) = lna + lnb to rewrite this equation as: lnNn1 + lnNn2 + n1lnn1 + n2lnn2 = ln(Nn1) + ln(Nn2) + ln(n1^n1) + ln(n2^n2)

Finally, we can use the fact that ln(ab) = ln(a) + ln(b) and ln(a^b) = blna to simplify the equation: ln(Nn1) + ln(Nn2) + ln(n1^n1) + ln(n2^n2) = ln(Nn1Nn2) + ln(n1^n1n2^n2)

Substituting back into the original equation, we get: NlnN - n1ln(n1) - n2ln(n2) = ln(Nn1Nn2) + ln(n1^n1n2^n2) = ln(Nn1Nn2) + ln((n1/N)^n1(n2/N)^n

 

[Architeuthis Dux]

To solve for NlnN - n1ln(n1) - n2ln(n2), we can use the fact that ln(a/b) = ln(a) - ln(b). Therefore, we can rewrite the expression as:

NlnN - n1ln(n1) - n2ln(n2) = NlnN - n1lnN + n1ln(n1/N) - n2lnN + n2ln(n2/N)

Now, we can regroup the terms as follows:

NlnN - n1lnN - n2lnN + n1ln(n1/N) + n2ln(n2/N)

Factoring out the common factor of N, we get:

N(lnN - lnN - lnN) + n1ln(n1/N) + n2ln(n2/N)

Simplifying further, we get:

N(0) + n1ln(n1/N) + n2ln(n2/N)

Which is equal to:

n1ln(n1/N) + n2ln(n2/N)

Finally, we can use the fact that ln(a/b) = -ln(b/a) to rewrite the expression as:

n1ln(n1/N) + n2ln(n2/N) = -n1ln(N/n1) - n2ln(N/n2)

And since N/n1 = n1/N and N/n2 = n2/N, we can rewrite the expression as:

-n1ln(n1/N) - n2ln(n2/N) = -n1ln(n1/N) - n2ln(n2/N)

Therefore, we have shown that NlnN - n1ln(n1) - n2ln(n2) is equal to -N{(n1/N)ln(n1/N) + (n2/N)ln(n2/N)}, by adding and subtracting terms.