$\begingroup$

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?

$\endgroup$ 1

2 Answers

$\begingroup$

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, $A^n\cdot B^n = (AB)^n$.

$\endgroup$ $\begingroup$

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:

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

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy