Just as a repeated sum $a + a + ... $k$ factors is written $a^k$. The number $k$ is called the exponent, and $a$ the base of the power $a^k$. https://www.britannica.com/science/arithmetic#ref24749 · In mathematics, exponentiation is an operation involving two numbers, the base ... More on math.stackexchange.com

The first definition of exponents is: a^n is a times itself n times. As you correctly pointed out, this only works when n is a natural number. But we can extend this definition by noticing the exponential property, which is that am+n = am ⋅ a^n. This is called the exponential property because it is true for exponents, and only for exponents; and it seems to encapsulate the very idea of exponentiation. Let me explain. The exponential property is true for all exponents. Suppose m and n are natural numbers. Then am+n is a times itself (m+n) times, which is a^m ⋅ a^n. This is only true for exponents. Suppose you have some function f such that f(m+n) = f(m)f(n) for all natural numbers m and n. Then by taking n=1+1+...+1, we see that f(n)=f(1)f(1)...f(1)=f(1)n, so f is an exponential function. The very idea of exponentiation is repeated multiplication. So an+1 is one more a than a^n, hence our rigorous version of the first definition, repeated multiplication, is of the form a1 = a and an+1=an ⋅ a. This formula is just an extension of that. Now, we can use this exponential formula to define all rational exponents. In particular, we get: Zero: We see that a^n = an+0 = a^n ⋅ a0. Therefore, a0 = 1. Negative integers: We see that 1 = a0 = an+(-n) = a^n ⋅ a-n. Therefore a-n = 1/(a^n). Fractions: consider the fraction m/n. We see that m/n+m/n+...+m/n = m if you add m/n to itself n times. So am/n⋅am/n⋅...⋅am/n = am/n+m/n+...+m/n = a^m. Therefore am/n is the n-th root of a^m. There we've extended the definition of exponentiation to all rational numbers. Now all that's left is the irrational reals (and maybe imaginary numbers, but that's for another day. The next thing we notice is that our new definition of exponents is continuous and always increasing for bases a>1 and always decreasing for 0 More on reddit.com

Another approach from the one mentioned below would be to first define e(x) by 1 + x + x²/2 + x³/6 + ... and then set yx to be e(ln(y)*x). Then, you can prove that this is continuous and coincides with the definition you know on the rationals. More on reddit.com

When you add a bunch of numbers, you implicitly start at 0. For instance if you add 2 3's together, you start with 0, then add 3, then add another 3. 0 is the natural starting point for addition. But if you multiply 2 3's together you don't implicitly start at 0. If you did you'd get 0 x 3 x 3 = 0. Instead you implicitly start at 1. 1 x 3 x 3 = 9. So 1 is the natural starting point for multiplication. Think of multiplication like scaling something (3 times as large, 4 times as large). Since 1 is the natural starting point for multiplication, it makes sense that when you multiply zero things together, you get 1. Just like when you add zero things together you get 0. More on reddit.com