ALGEBRA

Take the natural log of both sides
Ln1=-4/x+-10/y
Then use x=-4y/ln1+10


Geez, the ad banner isn't even joking, this is 10th grade mathematics

Taking the natural log of the entire equation.
ln(1)=ln(e^(-4/x))+ln(e^(-10/y))

ln(1) becomes, ln(e^n) becomes n
0=(-4/x)+(-10/y)

>>230706
>>230707
>ln(1)=ln(e^(-4/x))+ln(e^(-10/y))

That's not the same as:
ln((e^(-4/x))+(e^(-10/y)))

You can't just take the natural log of both terms individually, you have to take the natural log of the entire right side.

>>230729
>You can't just take the natural log of both terms individually, you have to take the natural log of the entire right side.

Sorry, I meant to end that with a question mark. I'm not telling you guys, I'm asking. I'm not sure.

>>230734
You're definitely right that you have to take the logarithm for the whole side and not just the terms individually. But I can't see how you would solve the problem. Hopefully another person can help.

>>230729
You can separate the terms
Ln(quantity)=ln(quan)*ln(tity)
So (-4/x)(-10/y)=0 works
Then use parametrics with R1/R2 and their respective rates to take a limit and define the relationship

Unfortunately, e^x+e^y=1 has an infinite number of solutions, so I don't have the time to completely solve this for you

I'm fairly certain you could use center of masses and a system for the first equation, and use the rate equations of those topics, but I don't have my mechanics book open

>>230741
>>230743
Thanks, guys.
This
>(-4/x)(-10/y)=0
should be enough for me to go off of for now.

>>230743
I would have to disagree with you there.
you can't just split a logarithm that way.
You can have ln(a*b) = ln(a) + ln(b), but not ln(a*b) = ln(a)*ln(b) as you suggest (or if you meant ln(a+b) = ln(a)*ln(b) that isn't valid either).
If I have misunderstood something be free to correct me.

>>230746
My bad, didn't see the title was algebra.
Hopefully you have success, this certainly is a strange problem for the topic

>>230751
Ln(x+y)=ln(quantity) I was using the definition of a quantity
e^x+e^y=c
Ln(e^x+e^y)=ln(c)
ln(e^[R1y/R1+y]+e^y)=ln(c)
([r1+y]{dr1/dy}+[r1y]{dr1+y}{r1+dy}+e^y)(ln[e^{R1y/y+R1}+e^y)/(e^R1y/R1+e^y)=1/c
Holding R2' constant

Then using limits, you can compare the relationships/convergences

Similar to finding pi with c/2r or limits of sin and cos and some other things I forget

I'm sure I made an error in the partial derivative, but im beat.
Unless I parametrized wrong in my head

>>230791
I apologize for my misunderstanding (didn't know quantity was a specific term), though I don't fully understand what you are doing (I wonder if the disappearing of x is a mistake or if I just don't understand what R1 and r1 are), but since I'm not OP you don't have to elaborate.

>>230807
No worries matey, I haven't been really "correct" since this is an algebra thread and I might have gone beyond that

Essentially, I'm not evaluating the function algebraically, instead using parametrization to model the separate rates, dy/dx and dR1/dR2
Using a partial derivative and limits to compare convergences between the parametrized functions isn't going to yield the same exact values but just model the relationship between them.

It looks wrong algebraically because it is, it is similar to l'hôpital's principle, the derivative of a limit is equal in convergence but not in value

I got carried away :^)