TITLE: I need "intuition" about fraction exponents, like $4^{1.2}$. What exactly is it meant to do with number $4$?
QUESTION [10 upvotes]: $4^3$ is $4\cdot 4\cdot 4$
Then $4^{1.2}$ is what, $4\cdot\dotso$??
What happens to the number with a fraction exponent when we try to represent it only using numbers and basic operations (like $+$, $-$, $\div$, and $\cdot$)?
$4^{1.2} = 4*4^{0.2}$; but there's still a fractional exponent, $0.2$, there. I can't represent $4^{1.2}$ only using basic operations.

Is there a way to do that? Is it impossible fundamentally in math, or is it possible using imaginary numbers?

i checked answers form this 8 year old question but it's not giving the answer i'm looking for.

REPLY [3 votes]: Seems intuitive that $p:=4^{1.2}$ should lie somewhere between $4^1=4$ and $4^2=16$. But where ?
To answer this in a consistent way only relying on integer powers, we make three remarks:

for any number $q$ and naturals $m,n$, $(q^m)^n=q^{mn}$ is always true;

it does not seem unreasonable to generalize to $m,n$ not integers;

accepting this rule, $p^5=(4^{1.2})^5=4^6=4096.$


Now we have to solve the equation
$$p^5=4096,$$ which we can do by trial and error.
For instance, we can write
$$4^5=1024,5^5=3125,6^5=7776,$$ and $p$ should lie between $5$ and $6$.
Then we refine by introducing decimals,
$$5^5=3125,5.1^5=3450.25251,5.2^5=3802.04032,5.3^5=4181.95493$$
and $p$ should lie between $5.2$ and $5.3$.
Continuing this search, we obtain (after very very lengthy computation)
$$p=5.2780316430915770374960078849186\cdots$$
It is possible to show that this number has infinitely many decimals, which never repeat periodically. It is said to be irrational. By means of this trick, we can in principle raise any number to a power that is a fraction. We can also raise to an irrational power by using a close rational approximation of that power.

In practice, computing arbitrary powers is made easy by means of the so-called logarithms and antilogarithms, using the formula
$$a^b=\text{antilog}(\log(a)\cdot b)$$
where $\log$ and $\text{antilog}$ denote two special functions that can be computed efficiently or tabulated. Observe that taking a power is now replaced by two function evaluations and a product.
(Historically, logarithms were introduced to ease the computation of products by hand, using $$a\cdot b=\text{antilog}(\log(a)+\log(b))$$ which trades a multiplication for an addition.)