Partial fractions on exponentials

[CuriousJ]

Is it possible to use partial fractions on exponential factors? An expression like this for example...

3^x
______________________
(2^x + 3^x)(5^x + 6^x)

Would it break down into something like this?
A^x/(2^x + 3^x) + B^x/(5^x + 6^x)

What are the rules for repeated factors and non-repeated, etc.? I am trying to figure out how the rules for partial fractions would apply to this type of problem. How would I solve for A and B?

 

[tiny-tim]

Hi CuriousJ!

(try using the X² tag just above the Reply box )

Is it possible to use partial fractions on exponential factors? An expression like this for example...

3^x
______________________
(2^x + 3^x)(5^x + 6^x)

Would it break down into something like this?
A^x/(2^x + 3^x) + B^x/(5^x + 6^x)

No, it's not possible …

you'd have an equation (I'll use f(x) and g(x) instead of your A^x and B^x) like f(x)(5^x + 6^x) + g(x)(2^x + 3^x) = 3^x, which isn't going to work.

 

[sparkster]

Could you use it if you made a substitution? If you had to integrate 1/e^x2+2e^x+1 you'd want to substitute u=e^x then use partial fractions.

 

[Mute]

Could you use it if you made a substitution? If you had to integrate 1/e^x2+2e^x+1 you'd want to substitute u=e^x then use partial fractions.

As long as you meant $1/(e^{2x} + 2e^x + 1)$ (that is, $e^{2x}$ as opposed to $e^{x^2}$, which is quite different), yes, I see no problems with using a "substitution" like that to partial fraction decompose it, for the particular case you've given.

The problem with the example in the OP is that you can't make a useful substitution like that in order to decompose the expression. You could write each term as u^{ln(b)}, where u = e^x and b is whatever the base of the exponential was, but the ln(b)'s won't be integers and so the decomposition would fail.

In order for the decomposition to work you'd need either all of the bases to be the same or integers powers of the lowest base.

 

[sparkster]

Right, e^2x. Thanks for catching that.

 

[CuriousJ]

The decomposition seems to work when the denominator is a conjugate pair and the numerator can be reduced to one of those bases...

Example:
2^x/((2^x + 5^x)(2^x - 5^x)) = 1/[2(5^x + 2^x)] - 1/[2(5^x - 2^x)]

Thanks tiny-tim and Mute, for answering my question.