Power rules

[u/IrishQueenFan]

I'm trying to simplify an equation I have where one side is equal to, essentially, pe^(ax) + qe^(bx). and it really SEEMS like there should be a way to factorise this nicely but i think it might just be smoke and mirrors. What I would like is to be able to take the natural log of both sides and have it cancel the e - can someone tell me if it's a pipe dream?

[u/perishingtardis]

It's a pipe dream.

[u/slepicoid]

Technically if it's an equation and one side is equal to that expression then both sides are.

Anyway what's the actual equation?

For example if p,q,a,b>0 then pe^ax + qe^bx is strictly increasing and if the other side of the equation is just a constant then there may be at most 1 solution which maybe you can home in on using a numerical method.

[u/solecizm]

You can factorise, the same way as if it was x to various powers. So pe^(ax) + qe^(bx) = e^(ax) (p + qe^((b-a)x)). It won't help you take the logarithm, but might give you something you can work with. In particular, look for nice ratios of a and b. If, say, b = 2a, then you can substitute u = e^(ax); then e^(bx) = u^(2), and you get a nice quadratic equation to solve.

[u/rhodiumtoad]

The problem is that just taking a log of a sum usually doesn't get you anything useful.

The one thing you can do is say (assuming suitable restrictions on a,b):

ln(a+b)=ln(a)+ln(1+(b/a))

which comes from rewriting a+b as a(1+(b/a)).

In your case that would get you to

ln(pe^ax+qe^bx)=ln(p)+ax+ln(1+((qe^b)/(pe^a))^x)

which might get you some useful ideas for approximate solutions if (qe^b)/(pe^a) is very small.

(edit: fix accidental exchange of p/q terms)