Multiplying numbers with different bases with the same exponent

Question

Solving a problem about simplification ("simplify $(2^n-2^{n-1})(3^n-3^{n-1})")$, I got this: $$ 2^n\cdot 3^n $$

I don't recall learning about multiplying numbers with different bases and same exponents, but doing some tests I guessed the answer was $6^n$, and it turned out to be correct.

Is this property always applicable? How can this property be proved in a generalized way?

Answer 1

Basically, it comes from the commutative and associative properties of multiplication. Let's say we had $2^3\cdot 3^3$. This gives $(2\cdot 2 \cdot 2)\cdot(3\cdot 3\cdot 3)$. Using associativity and commutativity, we can rewrite this as $(2\cdot 3)\cdot (2\cdot 3)\cdot (2\cdot 3)$ or, $6\cdot 6\cdot 6$ or $6^3$. As you can see, since the exponents are the same, we can match each base with the other base. It does not depend on the value of $n$. Therefore, $A^n\cdot B^n = (AB)^n$.

Answer 2

Let $a,b \in \mathbb{R}$ and $n \in \mathbb{N}$, basic way, we can interpret $a^n$ as the product of $a$ with itself $n$ times, so:

$(ab)^n=ab\cdots ab=a\cdots a b \cdots b=a^nb^n$.