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I have an expression $e^{x-a}(e^{bx}+c)$. I just want it multiplied throughout to get $e^{(b+1)x-a}+ce^{x-a}$. To achieve this I have tried Expand, Simplify, Collect (using pattern Exp[q_*x]). The best output I could get is

Expand[Exp[x-a]*(Exp[b*x]+c)]
(* Exp[-a+x+b*x]+c*Exp[-a+x] *)

I have tried Evaluate after this step to obtain proper exponent, but to no avail. How do I make it come with $e^{(b+1)x-a}$ in the first term?

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Make a rule:

Expand[Exp[x - a]*(Exp[b*x] + c)] /. Exp[a_] :> Exp[Collect[a, x]]

enter image description here

See also:

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MapAt with specified position:

tmp = Exp[x - a] (Exp[b x] + c) // Expand;
pos = Position[tmp, p_Plus /; ! FreeQ[p, x]];
MapAt[Collect[#, x] &, tmp, pos]    

c E^(-a + x) + E^(-a + (1 + b) x)

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  • 1
    $\begingroup$ +1 Thanks a lot. It is nice to be aware of different ways of approaching a problem. $\endgroup$ – Deep Nov 28 '16 at 4:48

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