proof of $\log(xy) =\log (x) + \log (y)$ $\log(xy) = \log (x) + \log ( y)$ and its division counter part 
were mentioned in an axiomatic way  which I failed to proof. 
I noted the correlation with the exponential rules of adding powers in case of multiplication of a single base and subtracting them in case of division but if we do proceed with that, how do we take log of terms linked with arithmetic operators ? 
it doesn't have to be base $10$ but I picked it for convenience.
another question , using the fact that y = log x to the base a is the inverse function of x = a^y does that mean that the identitiy mentioned at the begingin is the inverse of 10^x+y = 10^x times 10 ^ y ?
 A: $\log xy = \log x + \log y\\
10^{\log xy} = 10^{(\log x + \log y)}\\
10^{\log xy} = (10^{\log x})(10^{\log y})\\
xy = (x)(y)$
A: Let $r=\log(x), s=\log(y)$ and $t=\log(xy)$.
Then $x=10^r,y=10^s$ and $xy=10^t=10^r10^s=10^{r+s}$.
Therefore, $t=r+s$, that is, $\log(xy)=\log(x)+\log(y)$.
A: Let $b, x, y > 0$, with $b \neq 1$.  Suppose that $log_b x = u$ and that $\log_b y = v$.  Then, by definition of the logarithm, $b^u = x$ and $b^v = y$.  Therefore,
$$\log_b{xy} = \log_b(b^ub^v) = \log_b b^{u + v} = u + v = \log_b x + \log_b y$$
and 
$$\log_b\left(\frac{x}{y}\right) = \log_b \left(\frac{b^u}{b^v}\right) = \log_b b^{u - v} = u - v = \log_b x - \log_b y$$
A: I want to take a geometric view here. I promise it won't be too bad! Let's start with the fact that, with a base $b$,
$b^{n}b^{m} = b^{n+m}$
Which makes sense just from thinking about groupings of numbers,
$(b\cdot b \cdots b)_{ntot}\cdot(b \cdot b \cdots b)_{mtot} = (b\cdot b \cdots b)_{ntot+mtot}$.
Let's define a function $f(x)$ to be,
$f(x)=b^x$. 
Then we have the properties,
$f(n+m) = b^{n+m} = b^n b^m = f(n)f(m)$,
from what I discussed in the first sentence. But what does this actually mean? If we look at a graph (which I've crudely drawn because I am not good with drawing software),

This shows graphically what is happening with $n$ and $m$ and how the function responds (sorry for high quality?). But what about $log_b (x)$? Let's define a new function
$g(x) = log_b (x)$.
It turns out that $log_b(x)$ and $b^x$ are related by a reflection about the line $ y = x $ on a Cartesian grid. This is shown very nicely in this picture:
https://people.richland.edu/james/lecture/m116/logs/log2.gif
If you imagine the line $ y = x $ going straight through the origin you'll see that they are just reflections of each other! What does this have to do with your question? Well what if we imagine taking the $x$ axis in my original plot, and treating that as our new '$y$' axis (vertical axis), with the $f(x)$ axis being the $g(x)$ axis (horizontal axis). I've drawn another sketch (rotated by 90 degrees counterclockwise on right):

Now usually we like our positive direction to be to the right, so we'll flip it and I'll replot $n$ and $m$ on this graph:

Since it was just a reflection, we're seeing another exponentially increasing function! $n$ and $m$ are on our vertical 'x' axis and $g(n)$ and $g(m)$ are on our horizontal '$g(x)$' axis. But we know on the vertical axis the two values multiply to give a new value, then on the horizontal axis those two numbers add to give the final value for exponentially increasing functions! All we've done is swapped the axis and used the properties of $f(x)=b^x$ to show that,
$g(n)+g(m) = g(nm)$
For $g(x)=\log_b(x)$! Plugging in,
$\log_b(n) + \log_b(m) = \log_b(nm).$
A: Using a different definition of the logarithm, $\ln(x) := \int_1^x \frac{dt}{t}$, we can prove this without a definition of exponentiation.
$$
\ln(ab) = \int_1^{ab}\frac{dt}{t} = \int_1^a\frac{dt}{t} + \int_a^{ab} \frac{dt}{t}  = \ln(a) + \int_a^{ab} \frac{dt}{t}
$$
Next we can do a u-substitution with $u= \frac ta$, meaning the limits of the remaining integral can be re-expressed as $1$ and $b$. Also, note that $a du = dt$ as well as $au = t$, meaning we find that $ \int_a^{ab} \frac{dt}{t} = \int_1^b \frac{a\,du}{au} = \int_1^b \frac{du}{u} = \ln(b)$.
Now we find that $\ln(ab) = \ln(a) + \ln(b)$, using just basic properties of integrals.
A: Considering that $\log(x)$ is the inverse function of $e^x$ you have
$$e^{\log(xy)} = xy = e^{\log(x)} \cdot e^{\log(y)} = e^{\log(x)+\log(y)}$$
But since the exponential function is injective you get $\log(xy) = \log(x)+\log(y)$
