What are we doing when we raise $10$ to some decimal number? This is something I am having some difficulty to understand.
When I have $10^x$ where $x$ is some integer  greater than $0$ , I understand  that this is equivalent to write $$\underbrace{10\cdot 10 \cdot 10 \cdots 10}_{x  \text {times}} $$ 
but when I have the case that $10$ is raised to some decimal number  $x$ I really lack the intuition of what's going on here.
I ask this question because ,for example ,when I read that $10^{0,48}=3,01...$ I simply don't understand how we have that $10$ raised to some number equals $3$.In general why is it that every number can be described as $10^x$ ? 
P.s I  guess it's important that I say that I am yet in high school,so you know what kind of answers might be appropriate for me. 
 A: We have that $a^n$ for some natural number $n$ is equal to
$$
\underbrace{a\cdot a \cdots a}_{n\text{ times}}
$$
just as you said. This exponentiation operation fulfills a few relations, which I'm sure you've seen, such as $a^n\cdot a^m = a^{n+m}$, $(a^n)^m = a^{nm}$ and $a^n:a^m = a^{n-m}$ (but only for $n>m$ at the moment). As long as $n$ and $m$ are positive, this can be checked easily.
Say want to figure out what the best possible definition of $a^n$ is for negative integers $n$. Or $a^0$. What do we want from such a definition? Well, we want the same relations as above to apply. This, in specific, means that we want $1 = a^n:a^n = a^{n-n} = a^0$. The same rules also force $a^n\cdot a^{-n} = a^0 = 1$, which in turn implies $a^{-n} = \frac1{a^n}$. This paragraph, of course, requires that $a \neq 0$.
Now that we have defined $a^n$ for all integers, what about fractions? What would we want from a number such as $a^{1/n}$? Well, one thing we might want to demand is that $(a^{1/n})^n = a^{n/n} = a$. As long as $a> 0$, the equation $x^n = a$ has the single positive solution $x = \sqrt[n]{a}$, by definition of $n$-th root. Therefore, for the rule $(a^n)^m = a^{nm}$ to hold for rational powers $n, m$ and positive $a$, we must have $a^{1/n} = \sqrt[n]a$.
At this point, we must of course check that the definitions we have so far actually still fulfill the three rules I set up in the beginning. In every case, one of the rules forced a choice for a definition, but it could be that as we expanded the possible domain of $n$, the rules came at odds. It turns out they don't, but it has to be checked.
Now, for irrational exponents, there is no direct, satisfying definition that comes from expanding what we already have. We just require that the function $x \mapsto a^x$, which is previously defined for $x$ rational, is continuous for all real $x$. This forces a specific function value for any real number, and it turns out that the exponentiation rules still hold, so it's a good definition.
The alternative definition that people might use for real exponents is that
$
a^x = \operatorname{Exp}(x\ln a)
$where $$\operatorname{Exp}(x) = 1+x+\frac{x^2}2 +\frac{x^3}6 + \cdots + \frac{x^n}{n!} +\cdots$$
and $\ln a$ can be defined, for instance, by $\int_1^a \frac 1t dt$. This is a full definition that tells you exactly what $a^x$ is, but it doesn't pop nicely out of the original $a^n$-definition.
A: For the sake of illustration, I have made a plot of the powers of $2$ (those of $10$ grow too fast).

You see that they nicely align on a smooth and regular curve. Then it is a natural thing to define the powers for non-integer numbers too. And by continuity of the curve, for every number $y$, there will be some $x$ such that $y=2^x$.
How exactly we can compute these is another matter, but here is a hint: it is easy to show that $\sqrt{2^{2n}}=2^n$. Then, we can generalize to real $x$ by admitting $\sqrt{2^x}=2^{x/2}$. This tells us $2^{0.5}=\sqrt{2}, 2^{1.5}=2\sqrt2,2^{0.25}=\sqrt{\sqrt{2}},2^{0.75}=\sqrt{\sqrt{2}}\sqrt{2}\cdots$
A: By convention, if the power is a fraction of the form $\frac1n$ it means taking the $n$-th root. That is consistent with the rule, already true for integer exponents, that taking the $x-$th power of the $y$-th power is the same as taking the $(x.y)$-th power.
By extension then, if the power is any fraction, it means ordinary exponentiation with the power of the numerator of the fraction, followed by root-taking with the power of the denominator of the fraction:
$$a^{\frac{x}y}=\sqrt[y]{a^x}$$
If the exponent is irrational (i.e., a real number that is not a fraction) then the power can either be defined by a limit process (approximate the exponent every more closely with fractions). There is also a more sophisticated definition involving logarithms.
Thus, $10^{0.48}$ is the 12-th power of the 25-th root of 10, because 0.48 is equal to the fraction $\frac{12}{25}.$
A: For any finite decimals number $a=0.d_1d_2d_3\ldots d_n 000\ldots$ we have that $a=\frac{d_1d_2d_3\ldots d_n}{10^{n+1}}$.
When we write $x^a$ then what we mean is
$$x^a=(x^{d_1d_2d_3\ldots d_n})^{1/10^{n+1}}$$
That is we take x to the $d_1d_2d_3\ldots d_n$th power and then the $10^{n+1}$th root of it.
if $b$ is with infinite decimals (and presumingly not rational) we have
$$x^b=\lim_{n\to\infty}x^{a_i}$$
where $a_i$ are rational numbers that converges to $b$.
A: $$
\underbrace{10\cdot 10 \cdot 10 \cdots 10}_{x  \text {times}}
$$
is how $10^x$ is defined for natural numbers $x$. Mathematicians like to extend definitions, and that has happened with exponentiation too.
When doing these extensions, it's nice if they lead to pleasent results like well-known rules still applying.
A first step is observing that $a^m*a^n=a^{m+n}$, this allows us to see what raising to $\frac{1}{2}$ (and more generally any fraction) should be.
To understand the definition of raising to general real powers, you'll (probably - but someone might have a trick I don't know/remember) need to know logarithms.
As you now added a comment saying you know logarithms, I'll add a little.
The observation is that:
$$
a^q=\exp(q\log a)
$$
(using the same base for $\exp$ and $log$ of course, but any base will do), as that is a nice ($C^\infty$ for instance) function, that is used as the genral definition of exponentiation.
