Does anyone have the math skills to isolate the variable, a.

[VivaLaViking]

c is an arbitrary constant, a is a real number.

ln(a)
------- = c
(1 - a)

Isolate a.

[fishsauce]

[ QUOTE ]
c is an arbitrary constant, a is a real number.

ln(a)
------- = c
(1 - a)

Isolate a.

[/ QUOTE ]

The only equivalence I can find with only one occurence of a is

c = \int_1^a (x+1)/x dx

where \int_1^a is the definite integral from 1 to a. This is derived from the original equation by first multplying bth sides by (1-a), then exponentiating both sides to get
a e^a = e^(c+1).
Then taking the natural log of both sides you get
c+1 = ln(a e^a) = a + ln(a).
Since ln(a) = \int_1^a 1/x dx and a = \int_1^a 1 dx + 1,
we have
c+1 = \int_1^a (1 + 1/x) dx + 1.
But I think this is all you can do.

Also, from the second FTC, since c is constant,
0 = (a+1)/a
for which the only solution is a = -1, for which the original problem statement is undefined.

Anyhow, where does this problem come from?

[gumpzilla]

Answer: <font color="white"> Differentiate both sides, leading to 1 / (a * (1-a)) + ln a / (1 - a)^2 = 0. Now substitute in our relationship from before and manipulate to get a = - 1 / c. </font>

EDIT: I don't think this makes a lick of sense. At the very least the result I got doesn't make sense. It's time to go to bed.

[PairTheBoard]

[ QUOTE ]
fishsauce --

multplying bth sides by (1-a), then exponentiating both sides to get
a e^a = e^(c+1).


[/ QUOTE ]

You have an error here. You turned c(1-a) into c+1-a.

PairTheBoard

[PairTheBoard]

[ QUOTE ]
Answer: Differentiate both sides, leading to 1 / (a * (1-a)) + ln a / (1 - a)^2 = 0. Now substitute in our relationship from before and manipulate to get a = - 1 / c.
<font color="white">.
</font>
EDIT: I don't think this makes a lick of sense. At the very least the result I got doesn't make sense. It's time to go to bed.

[/ QUOTE ]

Taking the derivative makes no sense. On the left hand side you have a function in the variable a which is Not constant. Just because you're trying to solve for when the function equals c does not mean the derivative of the function is zero.

It would be like trying to solve the equation x^2=3 by saying the derivative of x^2 must be zero. ie. 2x=0 so x=0. It makes no sense.

PairTheBoard

[BluffTHIS!]

Doesn't the logarithmic base here need to be specified?

[Hiding]

ln, is the natural log (base 2)

[ QUOTE ]
ln, is the natural log (base 2)

[/ QUOTE ]
I assume you mean base e (2.71828.....)

[Hiding]

yep, sorry about that.

[VivaLaViking]

Can't anyone offer any suggestions on how variable, a, may be isolated.