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

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SUMMARY

The discussion centers on manipulating the expression eab, where 'a' is a constant and 'b' is a variable, to facilitate integration. The key insight is that the expression can be rewritten as (ea)b, allowing the constant 'a' to be factored out. The integral expression discussed is v3 / [ev(h/kT) - 1], which can be transformed into the form x3 / (ex - 1) through the substitution x = v(h/kT). This manipulation simplifies the integration process significantly.

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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.
 

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