Can I Separate a Constant and Variable Expression in Exponential Manipulation?

AI Thread Summary
The discussion focuses on separating a constant and variable in the expression e^ab for integration purposes. The user seeks to manipulate the expression to extract the variable b from an integral involving v, a variable, and (h/kT), a constant. It is clarified that the exponent rule x^m * x^n = x^(m+n) does not apply directly since the constant a is multiplied by the variable b in the same exponent. A suggested solution involves using the substitution x = v(h/kT), allowing the expression to be rewritten in a more manageable form. This approach effectively simplifies the integral for evaluation.
winterwind
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Homework Statement


I am working on a problem, and there is a small step I need help on:

I have the expression eab, where a is a constant, and b is a variable. I need to separate a and b so I can pull out the expression b from an integral expression. Is there any exponent law or manipulation I can do algebraically that will allow me to get separate the constant a expression from the variable b expression?

Homework Equations


Rules of exponents, algebraic manipulations.


The Attempt at a Solution


There isn't any simple rule (i.e. xmxn = xm+n) that I can use to separate the two expression. I've tried moving the expression around, to no avail. Any help is appreciated!
 
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winterwind said:

The Attempt at a Solution


(i.e. xmxn = xm+n) !

Why can't you use that? What is the integral expression you are talking about?
 
rock.freak667 said:
Why can't you use that? What is the integral expression you are talking about?

Unfortunately, that doesn't work, since the expression is eab. I would only be able to use that rule if it were eaeb. The constant a is multiplied to the variable b in the same exponential.

The integral expression is: v3/[ev(h/kT) -1] where v is the variable, and (h/kT) is constant. Essentially, I need to get the expression in the form x3/(ex-1) , where then I can use a table of integrals to evaluate the integral. Is there a simple way to get the above expression into that form? In this case, v = x, so I am close, but there is the (h/kT) constant term in the exponential that still needs to be taken care of.
 
Last edited:
You're overthinking it. Use the substitution x=\nu(h/kT).
 
eab = (ea)b and ea is just a constant.

Then ∫bx dx = bx/ln(b) + C
 
vela said:
You're overthinking it. Use the substitution x=\nu(h/kT).

Thanks! This is it.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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