Is it possible to directly factor out some exponent to some specific form and replace by some rule? For example I have


I want to use replacement

/. x^2 ->z

where I do NOT want to change the replacement rule like

% /. x^a_->z^(a/2) 

I wanted to know whether any simple procedure is there for direct replacement i.e.

x^4 /. {x^2->z} = z^2 
  • $\begingroup$ You could try x-> Sqrt[z]. $\endgroup$ – Henrik Schumacher Feb 21 '18 at 16:25
  • $\begingroup$ This is same as using x^a_->z^(a/2). As I said I do not want to change the replacement rule. The question is stupid, also Mathematica may not work like this. I wanted to know whether there is 'Something' for Something[x^4] /. {x^2->z} = z^2 $\endgroup$ – Boogeyman Feb 21 '18 at 16:28

I don't know if this helps for your actual use case, but you can try using Simplify:

Simplify[x^4, x^2==z]


  • $\begingroup$ Sorry this is not what I meant. This question has obviously many answer as way out. I was thinking of putting some module which factors out x^4 to x^2 * x^2 and do the replacement. $\endgroup$ – Boogeyman Feb 21 '18 at 17:00

It would be helpful if you could give more examples; but it sounds like what you're trying to do is replace a composite variable (x^2), rather than just x. You can do this by holding the expression containing the composite variable in an unevaluated state, as follows:

Unevaluated[x^2] /. x^2 -> z


For more complicated composite variables, you might need to do something like this:

Unevaluated[x^2] /. HoldPattern@x^2 -> z

That's not needed for the above example, but it is for this:

2 Exp[-\[Beta] (Subscript[\[Epsilon], 
   R] - \[Mu])] Exp[-\[Beta] \[Epsilon]]] /. 
   HoldPattern@Exp[-\[Beta] (Subscript[\[Epsilon], R] - \[Mu])] :> x

The one limitation you'll encounter is that the composite variable to be substituted needs to appear in the exact form you use in the replacement statement. Thus this will return z^2:

Unevaluated[x^2*x^2] /. x^2 -> z 

...but this will return x^4:

Unevaluated[x^4] /. x^2 -> z

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