2021-11-23 19:42:12 UTC, Juan Luis Varona:
>
> In the expression (5^x)^2-7*5^x+4, I want to substitute x^5 by t.
>
> With sagemath 7.2 (or another old versions), I can do
> ((5^x)^2-7*5^x+4).subs(5^x==t)
> and I get t^2 - 5*t + 4
>
> But sagemath 9.4 does not change the first 5^x and he gives
> 5^(2*x) - 7*t + 4
>
> Why?
>
> (In both cases, var("t") has been previously used)
>
> Yours,
> Juan Luis Varona
In recent versions of Sage, defining:
```
sage: t, x = SR.var('t, x')
sage: a = (5^x)^2-7*5^x+4
```
automatically groups exponents and gives:
```
sage: a
5^(2*x) - 7*5^x + 4
```
in which `5^x` is only seen once as such (old versions
of Sage possibly did not group exponents, thus keeping
two visible occurrences of `5^x` in the resulting `a`).
This means that only one `5^x` gets replaced
by `t` when we do the following substitution:
```
sage: aa = a.subs(5^x == t)
sage: aa
5^(2*x) - 7*t + 4
```
To work around this, we can instead think of
rewriting `x` as `log(t, 5)` as in the following
substitution, which gives the expected result:
```
sage: ab = a.subs(x == log(t, 5))
sage: ab
t^2 - 7*t + 4
```
Now we have a polynomial expression in t and
we can use corresponding tools. --Samuel
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