Why does $a^{b}·a^{c}=a^{b+c}$?

Question

Why does $a^{b}·a^{c}=a^{b+c}$ ?

I want proof for it, I asked professor and he replied: "I don't know, it's a property anyway that is true and that's all what you need to know."

Answer 1

If $b$ and $c$ are integers, then you just use the associative property for the multiplication. If $b$ and $c$ are real numbers, then you must request $a>0$. But also in that case the proof is hard and requires the definition of $a^b$ for real exponents $b$. This boils down to the completeness of the set $\mathbb{R}$ of real numbers. I tend to believe that this goes beyond your knowledge at the present time...

Answer 2

It is easy to see this using the definition of a power , if $b$ and $c$ are natural numbers.

$$a^b \cdot a ^c = \overbrace{a \cdots a}^{b\ times} \cdot \overbrace{a \cdots a}^{c\ times} = \overbrace{a \cdots a}^{b+c\ times} =a^{b+c}$$

With the rules

$$a^{-b}=\frac{1}{a^b}$$

and

$$a^{\frac{1}{b}}=\sqrt[b]{a}$$

it should be possible to extend it to the negative and rational numbers.

Answer 3

We can prove it using nothing but calculus and basic limits theorems, assuming you are defining $x^y$ as $x^y:=e^{\ln(x)\cdot y}$, where $e^x$ is the normal limit definition, and the sum property of $\ln$ following from calculus (see here).

Assume $y,z\in\mathbb{R}$ (although this argument is also true in $\mathbb{C}$, but would need some more precision) and that $x$ is nonzero. We want to show that $x^y\cdot x^z = x^{y+z}$. Let's work with the left-hand-side.

$$x^y\cdot x^z=e^{\ln(x)\cdot y}\cdot e^{\ln(x)\cdot z}= \lim_{n\rightarrow \infty}\left(1+\frac{\ln(x)\cdot y}{n}\right)^n\cdot\lim_{n\rightarrow \infty}\left(1+\frac{\ln(x)\cdot z}{n}\right)^n $$

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Since $x$, $y$ and $z$ are bounded, we know that $x^y$ and $x^z$ are too. These two limit terms at the end are bounded and exist.

Since the two limits are bounded, the product of the limits is the limit of the products (proofs of this are available in any intro. analysis text):

$$\lim_{n\rightarrow \infty}\left(1+\frac{\ln(x)\cdot y}{n}\right)^n\cdot\lim_{n\rightarrow \infty}\left(1+\frac{\ln(x)\cdot z}{n}\right)^n = $$

$$\lim_{n\rightarrow \infty}\left(1+\frac{\ln(x)\cdot y}{n}\right)^n\cdot\left(1+\frac{\ln(x)\cdot z}{n}\right)^n =$$

$$\lim_{n\rightarrow \infty}\left(\left[1+\frac{\ln(x)\cdot y}{n}\right] \cdot\left[1+\frac{\ln(x)\cdot z}{n}\right]\right)^n =$$

Compute:

$$\lim_{n\rightarrow \infty}\left(\left[1+\frac{\ln(x)\cdot (y+z)}{n}+\frac{\ln(x)^2\cdot y\cdot z}{n^2}\right]\right)^n =$$

$$e^{ \lim_{n\rightarrow \infty} n\cdot\ln{\left(1+\frac{\ln(x)\cdot (y+z)}{n}+\frac{\ln(x)^2\cdot y\cdot z}{n^2}\right)}}$$

Now lets think: if we let $k:=\frac{1}{n},$ then as $n$ goes to $\infty$, $k$ goes to $0$ s now we can rewrite this thing as a quotient:

$$=\exp \left[ \lim_{k\rightarrow 0} \frac{ \ln{\left(1+k\ln(x)\cdot (y+z) + k^2\ln(x)^2\cdot y\cdot z\right)}} {k} \right]$$

$\ln(x)^2\cdot y \cdot z$ and $\ln(x)\cdot (y+z)$ are both constants so the terms involving those go to 0. By continuity of $ln$, the numerator goes to $\ln(1)$ which is 0. The denominator also goes to 0. Now we can use L'hospitals rule (recall the standard proof, which isn't dependent on what we are trying to prove):

$$=\exp \left[ \lim_{k\rightarrow 0} \frac{\ln(x)(y+z)+2k\cdot \ln(x)^2\cdot y\cdot z} {\left(1+k\ln(x)\cdot (y+z) + k^2\ln(x)^2\cdot y\cdot z\right)} \right]$$

Now the limit of both the numerator and denominator exist and are nonzero so we can use the quotient rule of limits:

$$=\exp \left[ \frac{\lim_{k\rightarrow 0}\ln(x)(y+z)+2k\cdot \ln(x)^2\cdot y\cdot z} {\lim_{k\rightarrow 0}\left(1+k\ln(x)\cdot (y+z) + k^2\ln(x)^2\cdot y\cdot z\right)} \right]$$

And finally we can evaluate our limits:

$$=\exp \left[ \frac{\ln(x)(y+z)} {1} \right] = e^{\ln(x)(y+z)} = x^{y+z}$$

as expected.

Answer 4

HaHa, interesting!

Let us define an operation "$*$" on the set of positive integers $\mathbb{Z}_+\times \mathbb{Z}_+\to \mathbb{Z}_+$ as follows: $a*b=\underbrace{b\times b\times\cdots \times b}_{a\text{ times}}=b^a(\text{ usual exponentiation here })$. Define $a^2:= a*a$, inductively define $a^{n+1}:=a^{n}*a$. Then $a^3=a^2*a\neq a*a^2$ in general. For example, $3^2*3=\underbrace{3\times \cdots \times 3}_{27\text{ times}}$ but $3*3^2=27\times 27\times 27$. Therefore $3*3^2\neq 3^3$.

Just for fun!

Answer 5

It is quite simple. If you think properly you can see that: a^b= (axaxax...xa) {for b times} and simalrly for a^c we can have the same. Now,we are calculating the value of (a^b)x(a^c),you can say easily we are multiplying b numbers of a, and then we are multiplying the c numbers of a,with the value of the previous multiplication,silmple! Now the observation or the argument you can say,we are actually multiplying total (b+c) numbers of a's and very naturally if we assume(b+c=z)then we have z numbers of a's and if we multiply them we will get the value as a^z that is nothing but a^(b+c). so here we are proving the identity (a^b x a^c)=a^(b+c) by an simple observation. Hope the answer will help you!best of luck.

Answer 6

This is just bookkeeping. If you multiply $a^n$ and $a^m$, you have $m+n$ factors of $a$, so $a^{m+n} = a^m a^n$.