simplify (a^b)^c

[compsystems]

Hello

With what command can I simplify the following expression?


(3^x)^2/(2^x)^3 -> (9/8)^x

[parisse]

exp2pow(exp(lncollect(factor(lnexpand(ln((3^x)^2/(2^x)^3))))))

[compsystems]

Hola Bernard

As (a^b)^c = a^(b*c) = a^(c*b) = (a^c)^b

You can add a rule to the Xcas engine, that is, if b or c are numbers, place them before the variables, so that simplification can be given between numbers, this reduces the expression


(3^x)^2 = 3^(x*2) = 3^(2*x) => (3^2)^x = 9^x

[parisse]

Your rule requires additional assumptions. For example ((-1)^(2/3))^2 is not equal to ((-1)^2)^(2/3). And it is not always desirable to replace (3^x)^2 by 9^x.

[compsystems]

Your rule requires additional assumptions. For example ((-1)^(2/3))^2 is not equal to ((-1)^2)^(2/3).

What other assumptions should be taken?
It seems that the base must be a really positive number.
((1.5)^(2))^(2/3) -> 1.71707136383
((1.5)^(2/3))^2 -> 1.71707136383

And it is not always desirable to replace (3^x)^2 by 9^x.

In which cases?

[parisse]

For example solve 3^x+(3^x)^2=12.

[compsystems]

For example solve 3^x+(3^x)^2=12.

The solution is the same when exchanging the exponents (a^b)^c -> (a^c)^b

solve(3^x+(3^x)^2=12) -> x=1
solve(3^x+(3^2)^x=12) -> x=1
solve(3^x+(9)^x=12) -> x=1

a^(b^c) = a^b^c
a^(b^c) ≠ a^(c^d)
(a^b)^c = (a^c)^b = a^(b*c) = a^(c*b) = (a^c)^b

[parisse]

How do you think Xcas can solve 3^x+9^x=12?

[compsystems]

I do not know the computational algorithms of an algebraic system